## 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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