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.

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