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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