Co-ordinate geometry practice for IITJEE Maths: Ellipses

Problem 1:

Find the locus of the point of intersection of tangents to the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}}=1, which are at right angles.

Solution I:

Any tangent to the ellipse is y = mx + \sqrt{a^{2}m^{2}+b^{2}}….call this equation I.

Equation of the tangent perpendicular to this tangent is y=-\frac{-1}{m}x+\sqrt{\frac{a^{2}}{m^{2}}+b^{2}}…call this equation II.

The locus of the intersection of tangent lines (i) and (ii) is obtained by eliminating m between these equations. So, squaring and adding them, we get

(y-mx)^{2}+(my+x)^{2}=a^{2}m^{2}+b^{2}+a^{2}+b^{2}m^{2}

\Longrightarrow (1+m^{2})(x^{2}+y^{2})=(1+m^{2})(a^{2}+b^{2})

\Longrightarrow x^{2}+y^{2}=a^{2}+b^{2}

which is a circle with its centre at the centre of the ellipse and radius equal to the length of the line joining the ends of the major and minor axis. This circle is called the director circle of the ellipse.

Problem II:

A tangent to the ellipse x^{2}+4y^{2}=4 meets the ellipse x^{2}+2y^{2}=6 at P and Q. Prove that the tangents at P and Q of the ellipse x^{2}+2y^{2}=6 at right angles.

Solution II:

Let the tangent at R(2\cos {\theta}, \sin{\theta}) to the ellipse x^{2}+4y^{2}=4 meet the ellipse x^{2}+2y^{2}=6 at P and Q.

Let the tangents at P and Q to the second ellipse intersect at the point S(\alpha,\beta). Then, PQ is the chord of contact of the point S(\alpha,\beta) with respect to ellipse two, and so its equation is

\alpha x + 2\beta y=6….call this “A”.

PQ is also the tangent at R(2\cos {\theta}, \sin{\theta}) to the first ellipse and so the equation can be written as (2\cos{\theta})x+(4\sin{\theta})y=4….call this “B”.

Comparing “A” and “B”, we get \frac{2\cos{\theta}}{\alpha} = \frac{4\sin{\theta}}{2\beta} = \frac{4}{6}

\Longrightarrow \cos{\theta}=\frac{\alpha}{3} and \sin{\theta}=\frac{\beta}{3}

\Longrightarrow \frac{\alpha^{2}}{9} + \frac{\beta^{2}}{9}=1 \Longrightarrow \alpha^{2}+\beta^{2}=9

The locus of S( \alpha, \beta) is x^{2}+y^{2}=9, or x^{2}+y^{2}=6+3, which is the director circle of the second ellipse.

Hence, the tangents at P and Q to the ellipse (ii) are at right angles (by the solution to the previous example).

Problem 3:

Let d be the perpendicular distance from the centre of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1 to the tangent drawn at a point P on the ellipse. If F_{1} and F_{}{2} are the two foci of the ellipse, then show that (PF_{1}-PF_{2})^{2}=4a^{2}(1-\frac{b^{2}}{d^{2}})

Solution 3:

Equation of the tangent at the point P(a\cos {\theta}, b\sin{\theta}) on the given ellipse is \frac{x\cos{\theta}}{a} + \frac{y\sin{\theta}}{b}=1. Thus,

d= |\frac{-1}{\sqrt{\frac{\cos^{2}{\theta}}{a^{2}}+\frac{\sin^{2}{\theta}}{b^{2}}}}|

\Longrightarrow d^{2}=\frac{a^{2}b^{2}}{b^{2}\cos^{2}{\theta}+a^{2}\sin^{2}{\theta}}

We know PF_{1}+PF_{2}=2a

\Longrightarrow (PF_{1}-PF_{2})^{2}=(PF_{1}+PF_{2})^{2}-4PF_{1}PF_{2}…call this equation I.

Also, (PF_{1}.PF_{2})^{2}=[ (a\cos{\theta}-ae)^{2}+(b\sin{\theta})^ {2}].[(a\cos{\theta}+ae)^{2} + (b\sin{\theta})^{2}], which in turn equals,

[a^{2}(\cos{\theta}-e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ]. [a^{2}(\cos{\theta}+e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ], that is,

a^{4}[ (\cos^{2}{\theta}+e^{2}) -2e\cos{\theta}+\sin^{2}{\theta} - e^{2} \sin^{2}{\theta} ]. [ (\cos^{2}{\theta}+e^{2}) + 2e \cos{\theta} + \sin^{2}{\theta} - e^{2}\sin^{2}{\theta}] = a^{4}[ 1-2e\cos{\theta}+e^{2}\cos^{2}{\theta} ] [1+ 2e \cos{\theta} + e^{2}\cos^{2}{\theta} ]

that is,

a^{4}[ (1+e^{2}\cos^{2}{\theta})^{2}-4e^{2}\cos^{2}{\theta}] = a^{4}(1-e^{2}\cos^{2}{\theta})^{2}

\Longrightarrow PF_{1}. PF_{2}=a^{2}(1-e^{2}\cos^{2}{\theta})

Now, from I, we get (PF_{1} - PF_{2})^{2} = 4a^{2}-4a^{2}(1-e^{2}\cos^{2}{\theta}) = 4a^{2}e^{2}\cos^{2}{\theta},

also, 1-\frac{b^{2}}{a^{2}} = 1 - \frac{b^{2}\cos^{2}{\theta} + a^{2}\sin^{2}{\theta}}{a^{2}} = \frac{(a^{2}-b^{2})\cos^{2}{\theta}}{a^{2}} = e^{2}\cos^{2}{\theta}

Hence, (PF_{1} - PF_{2})^{2} = 4a^{2}(1-\frac{b^{2}}{d^{2}})

we will continue later,

Cheers,

Nalin Pithwa

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