## Monthly Archives: September 2017

### Conics: Co-ordinate Geometry for IITJEE Mains: Basics 5

Question I:

Given the base of a triangle and the ratio of the tangents of half the base angles, prove that the vertex moves on a hyperbola, whose foci are the extremities of the base.

Solution I:

Let the base of the triangle ABC be $BC=a$, and $\frac{\tan{(B/2)}}{\tan{C/2}}=K (constant)$

Then, we get $\Longrightarrow \sqrt{\frac{(s-c)(s-a)}{s(s-b)}} \times \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=K$

$\Longrightarrow \frac{(s-c)}{(s-b)}=K \Longrightarrow \frac{(a+b-c)}{(a-b+c)}=K$

$\Longrightarrow latex a+(b-c)=K(a-(b-c))$

$2K(b-c)=(K-1)a \Longrightarrow b-c=\frac{(K-1)a}{2K}$

Since a and K are given constants, $(K-1)a/2K$ is a constant, say $\alpha$. Therefore, we get $b-c=\alpha$, that is, $AB-AC=\alpha$. We find that the vertex A moves in such a manner that the difference of its distances from two fixed points B and C, which are the extremities of the base is constant. Hence, it describes a hyperbola whose foci are the extremities of the base. For instance, you will get further insight if you try the following: (in previous blog, basics 4, of coordinate geometry) : Show that the difference of focal distances of any point on a hyperbola is constant and equal to the length of the transverse axis.

Question 2:

Show that the angle between the tangents to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$ and the circle $x^{2}+y^{2}=ab$ at their points of intersection is $\arctan{\frac{(a-b)}{\sqrt{ab}}}$.

Solution 2:

For the points of intersection, we have $\frac{ab-y^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$

$\Longrightarrow y^{2}[\frac{1}{b^{2}} - \frac{1}{a^{2}}]=1-\frac{b}{a}$

$\Longrightarrow y^{2}=\frac{a^{2}b^{2}}{a^{2}-b^{2}}\times \frac{a-b}{a}=\frac{ab^{2}}{a+b}$

$y= \pm b \sqrt{\frac{a}{a+b}} \Longrightarrow \pm a \sqrt{\frac{b}{a+b}}$

Consider the point $P(\frac{a\sqrt{b}}{\sqrt{a+b}}, b\frac{\sqrt{a}}{a+b})$, intersection of the given ellipse and the circle. Equation of the tangent at P to the circle is $\frac{xa\sqrt{b}}{\sqrt{a+b}} + \frac{yb\sqrt{a}}{\sqrt{a+b}}=ab$

Slope of this tangent is $=-\frac{\sqrt{a}}{\sqrt{b}}$. Equation of the tangent at P to the ellipse is $\frac{xa\sqrt{b}}{a^{2}\sqrt{a+b}} + \frac{by\sqrt{a}}{b^{2}\sqrt{a+b}}=1$ and slope of this tangent is $=-\frac{b^{3/2}}{a^{3/2}}$. If $\alpha$ be the angle between these tangents, then $\tan{\alpha}=\frac{-\frac{b^{3/2}}{a^{3/2}}+ \frac{a^{1/2}}{b^{1/2}}}{1+ \frac{b^{3/2}.a^{1/2}}{a^{3/2}}.b^{-1/2}}=\frac{(a^{2}-b^{2})}{a^{1/2}.b^{1/2}(a+b)}=\frac{(a-b)}{\sqrt{ab}}$.

Hence, $\alpha = \arctan{\frac{(a-b)}{\sqrt{ab}}}$Note: The angle will be the same at each point of intersection.

Question 3:

A straight line is drawn parallel to the conjugate axis of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ to meet it and the conjugate hyperbola in the points P and Q respectively. Show that the normals at P and Q to the curves meet on the x-axis.

Solution 3:

Conjugate Hyperbola of the given hyperbola is the hyperbola whose transverse and conjugate axis are respectively the conjugate and transverse axis of the given hyperbola and thus the equation is $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$.

Now, $P(a\sec{\theta},b\tan{\theta})$ is a point of the given hyperbola and $Q(a\tan{\phi},b\sec{\phi})$ is a point on the conjugate hyperbola. Since P and Q lie on the line parallel to the conjugate axis $x=0$, of the given hyperbola, $a\sec{\theta}=a\tan{\phi}$, $\Longrightarrow \sec{\theta}=\tan{\phi}$.

Now, equation of the normal at P to the given hyperbola is $y-b\tan{\theta}=-\frac{a\tan{\theta}}{b\sec{\theta}}(x-a\sec{\theta})$…call this equation I.

and equation of the normal at Q to the conjugate hyperbola is $y-b\sec{\theta}=-\frac{a\sec{\phi}}{b\tan{\phi}}(x-a\tan{\phi})$….call this equation II.

Eliminating x from I and II, using $\sec {\theta}= \tan{\phi}$ we get: $\frac{b\sec{\theta}}{a\tan{\theta}}(y-b\tan{\theta})=\frac{b\tan{\phi}}{a\sec{\phi}}(y-b\sec{\phi})$, which in turn, implies that $y=0$. Hence, the normals meet on the x-axis.

More later, including homeworks,

Nalin Pithwa.

### Conics: Co-ordinate Geometry for IITJEE Mains: Problem solving: Basics 4

Question 1:

If the points of the intersection of the ellipses $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1$ and $\frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{\beta^{2}}=1$ are the end points of the conjugate diameters of the former, prove that :$\frac{a^{2}}{\alpha^{2}} + \frac{b^{2}}{\beta^{2}} = 2$

Solution 1:

The locus of the middle points of a system of parallel chords of an ellipse is a line passing through the centre of the ellipse. This is called the diameter of the ellipse and two diameters of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$ are said to be conjugate if each bisects the chords, parallel to the other. The condition for this is that the product of their slopes should be equal to $\frac{-b^{2}}{a^{2}}$.

Now, equation of the lines joining the centre $(0,0)$ to the points of intersection of the given ellipses is

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = \frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{\beta^{2}}$

$\Longrightarrow (\frac{1}{a^{2}}-\frac{1}{\alpha^{2}})x^{2}+(\frac{1}{b^{2}} - \frac{1}{\beta^{2}})y^{2}=0$…call this equation I;

If $m_{1}$, $m_{2}$ are the slopes of the lines represented by Equation I, then $m_{1}m_{2}=\frac{(\frac{1}{a^{2}}-\frac{1}{\alpha^{2}})}{(\frac{1}{b^{2}}-\frac{1}{\beta^{2}})}$

Since I represents a pair of conjugate diameters, $m_{1}m_{2}=-\frac{b^{2}}{a^{2}}$

Thus, $a^{2}(\frac{1}{a^{2}}-\frac{1}{\alpha^{2}})+b^{2}(\frac{1}{b^{2}}-\frac{1}{\beta^{2}})=0$

$\Longrightarrow \frac{a^{2}}{\alpha^{2}} + \frac{b^{2}}{\beta^{2}} =2$

Question 2:

Find the locus of the mid-points of the chords of the circle $x^{2}+y^{2}=16$ which are tangents to the hyperbola, $9x^{2}-16y^{2}=144$.

Solution 2:

Let $(h,k)$ be the middle point of a chord of the circle $x^{2}+y^{2}=16$.

Then, its equation is $hx + ky-16=h^{2}+k^{2}-16$, that is, $hx + ky=h^{2}+k^{2}$ ….call this equation I.

Let I touch the hyperbola: $9x^{2}-16y^{2}=144$

That is, $\frac{x^{2}}{16} - \frac{y^{2}}{9}=1$ …call this equation II.

at the point $(\alpha, \beta)$ say, then I is identical with

$\frac{x\alpha}{16} - \frac{y\beta}{9}=1$….call this equation III.

Thus, $\frac{\alpha}{16h} = \frac{-\beta}{9k} = \frac{1}{h^{2}+k^{2}}$

Since $(\alpha, \beta)$ lies on the hyperbola II,

$\frac{1}{16}(\frac{16h}{(h^{2}+k^{2})})^{2}-\frac{1}{9}(\frac{9k}{h^{2}+k^{2}})^{2}=1$

$\Longrightarrow 16h^{2}-9k^{2}=(h^{2}+k^{2})^{2}$.

Hence, the required locus of $(h,k)$ is $(x^{2}+y^{2})^{2}=16x^{2}-9y^{2}$.

Question 3:

If P be a point on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1$ whose ordinate is $y^{'}$, prove that the angle between the tangent at P and the focal chord through P is $\arctan{(\frac{b^{2}}{aey^{'}})}$.

Solution 3:

Let the coordinates of P be $(a\cos{\theta}, b\sin{\theta})$ so that $b\sin{\theta}=y^{'}$. Equation of the tangent at P is $\frac{x}{a}\cos{\theta}+\frac{y}{b}\sin{\theta}=1$.

Slope of the tangent is equal to $-\frac{b\cos{\theta}}{a\sin{\theta}}$

Slope of the focal chord SP is $\frac{b\sin{\theta}-0}{a\cos{\theta}-ae}=\frac{b\sin{\theta}}{a(\cos{\theta})-e}$.

If $\alpha$ is the required angle, then $\tan{\alpha}=\frac{-\frac{b\cos{\theta}}{a\sin{\theta}}-\frac{b\sin{\theta}}{a(\cos{\theta})-e}}{1-\frac{b\cos{\theta}b\sin{\theta}}{a\sin{\theta}a(\cos{\theta})-e}}$

which in turn equals $\frac{-b(\cos^{2}{\theta}-e\cos{\theta}+\sin^{2}{\theta})}{a\sin{\theta}(\cos{\theta}-e)} \times \frac{a^{2}(\cos{\theta}-e)}{(a^{2}-b^{2})\cos{\theta}-a^{2}e}$,

$=\frac{-ab(1-e\cos{\theta})}{\sin{\theta}(a^{2}e^{2}\cos{\theta}-a^{2}e)}=\frac{ab(e\cos{\theta}-1)}{a^{2}e\sin{\theta}(e\cos{\theta}-1)}$

$=\frac{ab}{a^{2}e\sin{\theta}}=\frac{ab.b}{a^{2}e.y^{'}}=\frac{b^{2}}{aey^{'}}$

$\Longrightarrow \alpha=\arctan{(\frac{b^{2}}{aey^{'}})}$

More later,

I hope you like it…my students should be inspired to try Math on their own…initially, it is slow, gradual, painstaking, but the initial “roots” pay very very “rich dividends” later…

-Nalin Pithwa.

### Conics: Co-ordinate Geometry: problem solving for IITJEE Mains: Basics 3

One more “nice” problem with solution ! Though, I strongly suggest you try the problem on your own and compare it with the solution given here. You might even come up with a clever approach than the one I try here:

Question:

If $r_{1}$, $r_{2}$ be the lengths of the radii vectors of the parabola $y^{2}=4ax$ which are drawn at right angles to one another from the vertex. Prove that $r_{1}^{4/3}r_{2}^{4/3}=16a^{2}(r_{1}^{2/3}+r_{2}^{2/3})$.

Solution:

Let $P(r_{1}\cos{\theta}, r_{2}\sin{\theta})$ and $Q(r_{2}\sin{\theta},r_{2}\cos{\theta})$ be two points on the parabola $y^{2}=4ax$ with lengths of the radii vectors $r_{1}$ and $r_{2}$ respectively, then $r_{1}^{2}\sin^{2}{\theta}=4ar_{1}\cos{\theta} \Longrightarrow r_{1}=\frac{4a\cos{\theta}}{\sin^{2}{\theta}}$

Similarly, $r_{2}=\frac{4a\sin{\theta}}{\cos^{2}{\theta}}$ since radii vectors $r_{1}$ and $r_{2}$ are at right angles.

Hence, $r_{1}r_{2}=\frac{16a^{2}}{\sin{\theta}\cos{\theta}}$

$\Longrightarrow (r_{1}r_{2})^{4/3}=\frac{(16a^{2})^{4/3}}{(\sin{\theta}\cos{\theta})^{4/3}}$…call this equation I.

And, $r_{1}^{2/3}+r_{2}^{2/3}=(4a)^{2/3}[\frac{\cos^{2/3}{\theta}}{\sin^{4/3}{\theta}} + \frac{\sin^{2/3}{\theta}}{\cos^{4/3}{\theta}}]$

which equals $\frac{(4a)^{2/3}}{\sin^{4/3}{\theta}\cos^{4/3}{\theta}}[\cos^{2}{\theta}+\sin^{2}{\theta}]$….call this equation II.

From I and II, we get : $\frac{r_{1}^{4/3}r_{2}^{4/3}}{r_{1}^{2/3}+r_{2}^{2/3}}=\frac{(16a^{2})^{4/3}}{(16a^{2})^{4/3}}=16a^{2}$.

More later,

Nalin Pithwa.