## 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.

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