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Category Archives: coordinate geometry
Coordinate geometry practice for IITJEE Maths: Ellipses
Problem 1:
Find the locus of the point of intersection of tangents to the ellipse , which are at right angles.
Solution I:
Any tangent to the ellipse is ….call this equation I.
Equation of the tangent perpendicular to this tangent is …call this equation II.
The locus of the intersection of tangent lines (i) and (ii) is obtained by eliminating m between these equations. So, squaring and adding them, we get
which is a circle with its centre at the centre of the ellipse and radius equal to the length of the line joining the ends of the major and minor axis. This circle is called the director circle of the ellipse.
Problem II:
A tangent to the ellipse meets the ellipse at P and Q. Prove that the tangents at P and Q of the ellipse at right angles.
Solution II:
Let the tangent at to the ellipse meet the ellipse at P and Q.
Let the tangents at P and Q to the second ellipse intersect at the point . Then, PQ is the chord of contact of the point with respect to ellipse two, and so its equation is
….call this “A”.
PQ is also the tangent at to the first ellipse and so the equation can be written as ….call this “B”.
Comparing “A” and “B”, we get
and
The locus of is , or , which is the director circle of the second ellipse.
Hence, the tangents at P and Q to the ellipse (ii) are at right angles (by the solution to the previous example).
Problem 3:
Let d be the perpendicular distance from the centre of the ellipse to the tangent drawn at a point P on the ellipse. If and are the two foci of the ellipse, then show that
Solution 3:
Equation of the tangent at the point on the given ellipse is . Thus,
We know
…call this equation I.
Also, , which in turn equals,
, that is,
that is,
Now, from I, we get ,
also,
Hence,
we will continue later,
Cheers,
Nalin Pithwa
Coordinate Geometry problems for IITJEE : equations of median, area of a triangle, and circles
Problem I:
If , and are the vertices of a triangle ABC, then prove that the equation of the median through A is given by:
Solution I:
If D is the midpoint of BC, its coordinates are
Therefore, equation of the median AD is , which in turn, implies that,
Now apply the row transformation to the previous determinant. So, we get
, using the sum property of determinants.
Hence, the proof.
Problem 2:
If is the area of the triangle with vertices , and is the area of the triangle with vertices , , and , and is the area of the triangle with vertices , , . Then, prove that there is no value of for which the areas of triangles, , and are in GP.
Solution 2:
We have , and
.
Applying the following column transformations to the above determinant, and , we get
and
so that .
Now, , and are in GP, if
, that is,
, where . But, for this value of , the vertices of the given triangles are not defined. Hence, , and and cannot be in GP for any value of .
Problem 3:
Two points P and Q are taken on the line joining the points and such that . Circles are drawn on AP, PQ, and QB as diameters. The locus of the point S, the sum of the squares of the length of the tangents from which to the three circles is equal to , is
(a)
(b)
(c)
(d) .
Ans. b.
Solution 3:
Since , the coordinates of P are and of Q are , equations of the circles on AP, PQ, and QB as diameters are respectively.
Please draw the diagram.
So, we get
So, if be any point of the locus, then .
So, the required locus of is .
More later,
Nalin Pithwa.
Coordinate Geometry : IITJEE Mains practice: some random problems again
Problem 1:
The line cuts the circle at P and Q. The line cuts the circle at R and S. If P, Q, R and S are concyclic, prove that
.
Solution I;
An equation of a circle through P and Q is …call this equation I.
And, an equation of a circle through R and S is …call this equation II.
If P, Q, R and S are concyclic, then I and II represent the same circle for same values of and .
or
so also,
or
or .
Eliminating and , we get the following:
, that is,
Problem II:
A straight line is such that the algebraic sum of the perpendiculars falling upon it from any number of fixed points is zero. Show that it always passes through a fixed point.
Solution II:
Let where be n fixed points. Let be the given line. Thus, as per given hypthesis, we have
where and
which shows that the given line passes through the fixed point .
Problem III:
The straight lines and are intersecting. Find the straight line such that L is the bisector of the angle between and .
Solution III:
Let the equation of the line be where the slopes of are respectively
.
Since L is the bisector of the angle between and we have
Hence, the equation of the required line is .
Problem IV:
If a, b are real numbers and , find the locus represented by .
PS: Please draw a right angled triangle PMA, with right angle at M, and P being and A being .
Solution IV:
Let and , then the given equation becomes .
where and which is the slope of , which in turn implies
, or . The given equation now becomes
….call this as relation I.
If M is the foot of the perpendicular from a point P(x,y) on the line and A is the point which clearly lies on this line, then from relation I, we have
. Hence, the locus of P is a straight line through the point inclined at an angle with the line .
Problem V:
Find the coordinates of the orthocentre of the triangle formed by the lines and and , where , and show that for all values of t and u, the orthocentre lies on the line .
Solution V:
Let the equation of the side BC be . Then, the coordinates of B and C are and , respectively, where and are equations of AB and AC, respectively.
PS: Please draw the diagram on your own for a better understanding of the solution presented.
Now, equation of BE is …let us call this equaiton I.
And, equation of CF is …let us call this equation II.
Solving I and II, we get the following:
, which in turn implies that
and , so that the orthocentre is the point which lies on the line .
Cheers,
Nalin Pithwa
Some random problems/solutions in Coordinate Geometry II: IITJEE mathematics training
Question I:
Find the equation of the tangent to the circle at the point . If the circle rolls up along this tangent by 2 units, find its equation in the new position.
Solution I:
The centre of the given circle is and its radius is 2. Equation of the tangent at to the circle is
or .
The slope of this line is showing that it makes an angle of 60 degrees with the xaxis. After the circle rolls up along the tangent at A through a distance 2 units, its centre moves from to . We now find the coordinates of . Since is parallel to the tangent at A and it passes through , , its equation is , where ; being at a distance 2 units on this line from ; its coordinates are
, that is, .
Hence, the equation of the circle in the new position is
, which in turn implies that
.
Question 2:
A triangle has two of its sides along the axes, its third side touches the circle . Prove that the locus of the circumcentre of the triangle is
.
Solution 2:
The given circle has its centre at and its radius is a so that it touches both the axes along which lie the two sides of the triangle. Let the third side be .
So that A is and B is and the line AB touches the given circle. Since is a right angle, AB is diameter of the circumcentre of the triangle AOB. So, the circumcentre of the triangle AOB is the midpoint of AB,
that is, , .
Now, the equation of AB is , which touches the given circle,
.
Hence, the locus of is .
Question 3:
A circle of radius 2 units rolls on the outerside of the circle , touching it externally. Find the locus of the centre of this outside circle. Also, find the equations of the common tangents of these two circles when the line joining the centres of the two circles make an angle of 60 degrees with xaxis.
Solution 3:
The centre C of the given circle is and its radius is 2. Let be the centre of the outer circle touching the given circle externally then , which in turn implies,
So, the locus of P is , or .
Since the two circles touch each other externally,, there are 3 common tangents to these circles.
One will be perpendicular to the line joining the centres and the other two will be parallel to the line joining the centres as the radii of the two circles are equal, coordinates of P are given by
,
coordinates of M, the midpoint of CP is .
Hence, the equation of the common tangent perpendicular to CP is
.
Let the equation of the common tangent parallel to CP be .
Since it touches the given circle .
Hence, the other common tangents are .
Question 4:
If and are the equations of two circles with radii r and respectively, then show that the circles cut orthogonally.
Solution 4:
Let the line of centres of the given circle be taken as the xaxis and its midpoint as the origin…Note this is the key simplifying assumption.
If the distance between the centres is 2a, the coordinates of the centre are and . Hence, we get the following:
, that is,
and so that , that is,
…call this I.
and and in turn …call this II.
Now, since .
The circles I and II intersect orthogonally.
Question 5:
Let P, Q, R, S be the centres of the four circles each of which is cut by a fixed circle orthogonally. If , , , be the squares of the lengths of the tangents to the four circles from a point in their plane, then prove that
Solution 5:
Let the equations of the four circles be
, , then centres of these circles are as follows:
, , , and
Let the fixed point in the plane be taken as the origin, then , , and . Let the equation of the fixed circle cutting the four circles orthogonally be
, then , or
we get the following:
, for .
Eliminating the unknowns g, f, c we get
or,
where ,
and , and
Hence, we get the following:
.
Homework Quiz Coordinate Geometry:
 OAB is any chord of a circle which passes through O, a point in the plane of the circle and meets it in points A and B. A point P is taken on this chord such that OP is (i) arithmetic mean (ii) geometric mean of OA and OB. Prove that the locus of P in either case is a circle. Determine the circle.
 Let be the equation of a pair of tangents drawn from the origin O to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, find the length OA.
 Let P, Q and R be the centres and are the radii respectively of three coaxial circles. Show that
 If ABC be any triangle and be the triangle formed by the polars of the points A, B, C with respect to a circle, so that is the polar of A; is the polar of B and is the polar of C. Prove that the lines , and meet in a point.
That’s all, folks !
Nalin Pithwa.
Some random sample problemssolutions in Coordinate Geometry: I: IITJEE Mains Maths tutorials
Question I:
The point undergoes the following transformations, successively:
a) reflection about the line .
b) translation through a distance 2 units along the positive directions of the xaxis.
c) rotation through an angle of about the origin in the anticlockwise direction.
d) reflection about .
Hint: draw the diagrams at very step!
Ans:
Question 2:
are n points in a plane whose coordinates are , , , respectively. , is bisected at the point , is divided in the ratio 1:2 at , is divided in the ratio at , is divided in the ratio at and so on until all n points are exhausted. Show that the coordinates of the final point so obtained are
.
Solution 2:
The coordinates of are .
Now, divides in the ratio . Hence, the coordinates of are
, or .
Again, divides in the ratio . Therefore, the coordinates of are , or
.
Proceeding in this manner,we can show that the coordinates of the final point obtained will be
.
Remark: For a rigorous proof, prove the above by mathematical induction.
Question 3:
A line L intersects the three sides BC, CA, and AB of a triangle ABC at P, Q and R, respectively. Show that
Solution 3:
Let , , and be the vertices of , and let be equation of the line L. If P divides BC in the ratio , then the coordinates of P are .
Also, as P lies on L, we have
…..call this relation I.
Similarly, we can obtain ….call this relation II.
and so, also, we can prove that …call this III.
Multiplying, I, II and III, we get the desired result.
The above is the famous Menelaus’s theorem of plane geometry proved with elementary tools of coordinate geometry. As a homework quiz, try proving the equally famous Ceva’s theorem of plane geometry with elementary tools of coordinate geometry.
Question 4:
A triangle has the lines and as two of its sides, with and being roots of the equation . If is the orthocentre of the triangle, show that the equation of the third side is .
Solution 4:
Since the given lines intersect at the origin, one of the triangle lies at the origin O(0,0). Let OA and OB be the given lines and , respectively. Let the equation of AB be . Now, as OH is perpendicular to AB, we have
, , say…call this equation I
Also, the coordinates of A and B are respectively,
and
Therefore, the equation of AB is
or …call this II.
Similarly, the equation of BH is ….call this III.
Solving II and III, we get the coordinates of H. Subtracting III from II, we get
Since and are the roots of the equation , we have and .
because y=b for H.
.
Hence, the equation of AB is
More later,
Nalin Pithwa.
Circles and System of Circles: IITJEE Mains: some solved problems I
Part I: Multiple Choice Questions:
Example 1:
Locus of the midpoints of the chords of the circle which subtend a right angle at the centre is (a) (b) (c) (d)
Answer 1: C.
Solution 1:
Let O be the centre of the circle , and let AB be any chord of this circle, so that . Let be the midpoint of AB. Then, OM is perpendicular to AB. Hence, . Therefore, the locus of is .
Example 2:
If the equation of one tangent to the circle with centre at from the origin is , then the equation of the other tangent through the origin is (a) (b) (c) (d) .
Answer 2: C.
Solution 2:
Since touches the given circle, its radius equals the length of the perpendicular from the centre to the line . That is,
.
Let be the equation of the other tangent to the circle from the origin. Then,
, which gives two values of m and hence, the slopes of two tangents from the origin, with the product of the slopes being 1. Since the slope of the given tangent is 3, that of the required tangent is 1/3, and hence, its equation is .
Example 3.
A variable chord is drawn through the origin to the circle . The locus of the centre of the circle drawn on this chord as diameter is (a) (b) (c) (d) .
Answer c.
Solution 3:
Let be the centre of the required circle. Then, being the midpoint of the chord of the given circle, its equation is .
Since it passes through the origin, we have .
Hence, locus of is .
Quiz problem:
A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If m and n are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is (a) (b) (c) (d) .
To be continued,
Nalin Pithwa.
Circles and Systems of Circles: IITJEE mains coordinate geometry basics
Section I:
Definition of a Circle:
A circle is the locus of a point which moves in a plane so that it’s distance from a fixed point in the plane is always constant.The fixed point is called the centre of the circle and the constant distance is called its radius.
Section II:
Equations of a circle:
 An equation of a circle with centre and radius r is .
 An equation of a circle with centre and radius r is .
 An equation of a circle on the line segment joining and as diameter is .
 General equation of a circle is : where g, f, and c are constants
 centre of this circle is
 Its radius is ,
 Length of the intercept made by this circle on the xaxis is if and that on the yaxis is if .
 General equation of second order degree in x and y represents a circle if and only if:
 coefficient of equals coefficient of , that is,
 coefficient of is zero, that is ,
Section III: Some results regarding circles:
 Position of a point with respect to a circle: Point lies outside, on or inside a circle , according as
 Parametric coordinates of any point on the circle are given by with . In particular, parametric coordinates of any point on the circle.
 An equation of the tangent to the circle at the point on the circle is
 An equation of the normal to the circle at the point on the circle is
 Equations of the tangent and normal to the circle at the point on the circle are, respectively, and
 The line is a tangent to the circle if and only if .
 The lines are tangents to the circle , for all finite values of m. If m is infinite, the tangents are .
 An equation of the chord of the circle , whose midpoint is is , where and . In particular, an equation of the chord of the circle , whose midpoint is is .
 An equation of the chord of contact of the tangents drawn from a point outside the circle is .(S and T are as defined in (8) above).
 Length of the tangent drawn from a point outside the circle , to the circle, is . (S and ) are as defined in (8) above.)
 Two circles with centres and and radii , respectively, (i) touch each other externally if . the point of contact is and (ii) touch each other internally if , where ; the point of contact is
 An equation of the family of circles passing through the points and is , where
 An equation of the family of circles which touch the line at for any finite value of m is . If m is infinite, the equation becomes .
 Let QR be a chord of a circle passing through the point and let the tangents at the extremities Q and R of this chord intersect at the point . Then, locus of L is called the polar of P with respect to the circle, and P is called the pole of its polar.
 Equation of the polar of with respect to the circle is , where T is defined as above.
 If the polar of P with respect to a circle passes through Q, then the polar of Q with respect to the same circle passes through P. Two such points P and Q, are called conjugate points of the same circle.
 If lengths of the tangents drawn from a point P to the two circles and are equal, then the locus of P is called the radical axis of the two circles and , and its equation is , that is,
 Radical axis of two circles is perpendicular to the line joining their circles.
 Radical axes of three circles, taken in pairs, pass through a fixed point called the radical centre of the three circles, if the centres of these circles are noncollinear.
4: Special Forms of Equation of a Circle:
 An equation of a circle with centre and radius is . This touches the coordinate axes at the points and .
 An equation of a circle with centre , radius is . This touches the xaxis at .
 An equation of a circle with centre and radius is . This circle passes through the origin , and has intercepts a and b on the x and y axes, respectively.
5: Systems of Circles:
Let ; and and .
 If two circles and intersect at real and distinct points, then where represents a family of circles passing through these points, where is a parameter, and when represents the chord of the circles.
 If two circles and touch each other, then represents equation of the common tangent to the two circles at their point of contact.
 If two circles and intersect each other orthogonally (the tangents at the point of intersection of the two circles are at right angles), then .
 If the circle intersects the line at two real and distinct points, then represents a family of circles passing through these points.
 If is a tangent to the circle at P, then represents a family of circles touching at P, and having as the common tangent at P.
 Coaxial Circles: A system of circles is said to be coaxial if every pair of circles of the system have the same radical axis. The simplest form of the equation of a coaxial system of circles is : , where g is a variable and c is constant, the common radical axis of the system being yaxis and the line of centres being xaxis. The Limiting points of the coaxial system of circles are the members of the system which are of zero radius. Thus, the limiting points of the coaxial system of circles are if . The equation () represents a family of coaxial circles, two of whose members are given to be and .
 Conjugate systems (or orthogonal systems) of circles : Two system of circles such that every circle of one system cuts every circle of the other system orthogonally are said to be conjugate system of circles. For instance, and , where g and f are variables and c is constant, represent two systems of coaxial circles which are conjugate.
6: Common tangents to two circles:
If and are two circles with centres and and radii and respectively, then we have the following results regarding their common tangents:
 When , that is, distance between the centres is greater than the sum of their radii, the two circles do not intersect with each other, and four common tangents can be drawn to circles. Two of them are direct common tangents and other two are transverse common tangents. The points of intersection of direct common tangents and transverse common tangents respectively, always lie on the line joining the centres of the two circles and divide it externally and internally respectively in the ratio of their radii.
 When , that is, the distance between the centres is equal to the sum of their radii, the two circles touch each other externally, two direct tangents are real and distinct and the transverse tangents coincide.
 When , that is, the distance between the centres is less than the sum of the radii, the circles intersect at two real and distinct points, the two direct common tangents are real and distinct while the transverse common tangents are imaginary.
 When with , that is, the distance between the centres is equal to the difference of their radii, the circles touch each other internally, two direct common tangents are real and coincident, while the transeverse common tangents are imaginary.
 When , with , that is, the distance between the centres is less than the difference of the radii, one circle with smaller radius lies inside the other and the four common tangents are all imaginary.
To be continued,
Nalin Pithwa.
Cartesian System, Straight Lines: IITJEE Mains: Problem Solving Skills II
I have a collection of some “random”, yet what I call ‘beautiful” questions in Coordinate Geometry. I hope kids preparing for IITJEE Mains or KVPY or ISI Entrance Examination will also like them.
Problem 1:
Given n straight lines and a fixed point O, a straight line is drawn through O meeting lines in the points , , , , and on it a point R is taken such that
Show that the locus of R is a straight line.
Solution 1:
Let equations of the given lines be , , and the point O be the origin .
Then, the equation of the line through O can be written as where is the angle made by the line with the positive direction of xaxis and r is the distance of any point on the line from the origin O.
Let be the distances of the points from O which in turn and , where .
Then, coordinates of R are and of are where .
Since lies on , we can say for
, for
…as given…
Hence, the locus of R is which is a straight line.
Problem 2:
Determine all values of for which the point lies inside the triangle formed by the lines , , .
Solution 2:
Solving equations of the lines two at a time, we get the vertices of the given triangle as: , and .
So, AB is the line , AC is the line and BC is the line
Let be a point inside the triangle ABC. (please do draw it on a sheet of paper, if u want to understand this solution further.) Since A and P lie on the same side of the line , both and must have the same sign.
or which in turn which in turn either or ….call this relation I.
Again, since B and P lie on the same side of the line , and have the same sign.
and , that is, …call this relation II.
Lastly, since C and P lie on the same side of the line , we have and have the same sign.
that is
or ….call this relation III.
Now, relations I, II and III hold simultaneously if or .
Problem 3:
A variable straight line of slope 4 intersects the hyperbola at two points. Find the locus of the point which divides the line segment between these two points in the ratio .
Solution 3:
Let equation of the line be where c is a parameter. It intersects the hyperbola at two points, for which , that is, .
Let and be the roots of the equation. Then, and . If A and B are the points of intersection of the line and the hyperbola, then the coordinates of A are and that of B are .
Let be the point which divides AB in the ratio , then and , that is, …call this equation I.
and ….call this equation II.
Adding I and II, we get , that is,
….call this equation III.
Subtracting II from I, we get
so that the locus of is
More later,
Nalin Pithwa.
Cartesian system and straight lines: IITJEE Mains: Problem solving skills
Problem 1:
The line joining and is produced to the point so that , then find the value of .
Solution 1:
As M divides AB externally in the ratio , we have and which in turn
Problem 2:
If the circumcentre of a triangle lies at the origin and the centroid in the middle point of the line joining the points and , then where does the orthocentre lie?
Solution 2:
From plane geometry, we know that the circumcentre, centroid and orthocentre of a triangle lie on a line. So, the orthocentre of the triangle lies on the line joining the circumcentre and the centroid , that is, , or . That is, the orthocentre lies on this line.
Problem 3:
If a, b, c are unequal and different from 1 such that the points , and are collinear, then which of the following option is true?
a:
b:
c:
d:
Solution 3:
Suppose the given points lie on the line then a, b, c are the roots of the equation :
, or
and , that is,
Eliminating l, m, n, we get
, that is, option (d) is the answer.
Problem 4:
If and are in A.P., with common difference a and b respectively, then on which line does the centre of mean position of the points with lie?
Solution 4:
Note: Centre of Mean Position is .
Let the coordinates of the centre of mean position of the points , be then
and
,
and
, and
, that is, the CM lies on this line.
Problem 5:
The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then what is the value of ?
Solution 5:
Equation of the line L in the two coordinate systems is , and where are the new coordinate of a point when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed.
or . So, the value is zero.
Problem 6:
Let O be the origin, and and are points such that and , then which of the following options is true:
a: P lies either inside the triangle OAB or in the third quadrant
b: P cannot lie inside the triangle OAB
c: P lies inside the triangle OAB
d: P lies in the first quadrant only.
Solution 6:
Since , P either lies in the first quadrant or in the third quadrant. The inequality represents all points below the line . So that and imply that either P lies inside the triangle OAB or in the third quadrant.
Problem 7:
An equation of a line through the point whose distance from the point has the greatest value is :
option i:
option ii:
option iii:
option iv:
Solution 7:
Let the equation of the line through be . If p denotes the length of the perpendicular from on this line, then
, say
then is greatest if and only if s is greatest.
Now,
so that . Also, , if , and
, if
and , if . So s is greatest for . And, thus, the equation of the required line is .
Problem 8:
The points , , Slatex C(4,0)$ and are the vertices of a :
option a: parallelogram
option b: rectangle
option c: rhombus
option d: square.
Note: more than one option may be right. Please mark all that are right.
Solution 8:
Midpoint of AC =
Midpoint of BD =
the diagonals AC and BD bisect each other.
ABCD is a parallelogram.
Next, and and since the diagonals are also equal, it is a rectangle.
As and , the adjacent sides are not equal and hence, it is neither a rhombus nor a square.
Problem 9:
Equations and will represent the same line if
option i:
option ii:
option iii:
option iv:
Solution 9:
The two lines will be identical if there exists some real number k, such that
, and , and .
or
or , and
or
That is, or , or .
Next, . Hence, , or .
Problem 10:
The circumcentre of a triangle with vertices , and lies at the origin, where and . Show that it’s orthocentre lies on the line
Solution 10:
As the circumcentre of the triangle is at the origin O, we have , where r is the radius of the circumcircle.
Hence,
Therefore, the coordinates of A are . Similarly, the coordinates of B are and those of C are . Thus, the coordinates of the centroid G of are
.
Now, if is the orthocentre of , then from geometry, the circumcentre, centroid, and the orthocentre of a triangle lie on a line, and the slope of OG equals the slope of OP.
Hence,
because .
Hence, the orthocentre lies on the line
.
Hope this gives an assorted flavour. More stuff later,
Nalin Pithwa.