Question

The tangents drawn from a point P to the ellipse makes an angle ${{\theta }_{1}}\text{ and }{{\theta }_{2}}$with the major axis. Find the locus of P when:
${{\tan }^{2}}{{\theta }_{1}}+{{\tan }^{2}}{{\theta }_{2}}=\lambda \left( \text{constant} \right)$

Answer 1

In the above question it is asked to find the locus of a point P which lies at the intersecting tangents of the ellipse. The tangents that touch the ellipse are the endpoints of the major axis. For this to work out we are first going to assume the x and y coordinate of P, then we will be using the tangent equation of ellipse and then substitute the assumed x and y coordinates to get a quadratic equation with use of quadratic equation we can simply use the above relation which is given as : ${{\tan }^{2}}{{\theta }_{1}}+{{\tan }^{2}}{{\theta }_{2}}=\lambda \left( \text{constant} \right)$

Complete step by step solution:
We know the equation of ellipse which is stated as under:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
So, for first we are going to assume the x and y coordinates of P as h and k and we got:
$\Rightarrow p\left( x,y \right)\text{ }=\text{ }p\left( h,k \right)$
Now we also know the tangent equation for ellipse which is stated as below:
$y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}$
We know that this is the equation of tangent and point p lies on the tangent. We can substitute the x and y coordinates i.e. h and k, in place of x and y in the equation and we get.

& \Rightarrow k=mh\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}} \\\ & \Rightarrow k-mh=\pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}} \\\ \end{aligned}$$ Now we will be squaring both the sides, from which we will get, $$\begin{aligned} & \Rightarrow {{\left( k-mh \right)}^{2}}={{\left( \pm \sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}} \right)}^{2}} \\\ & \Rightarrow \left( {{k}^{2}}+{{m}^{2}}{{h}^{2}}-2mkh \right)={{a}^{2}}{{m}^{2}}+{{b}^{2}} \\\ & \Rightarrow \left( {{h}^{2}}-{{a}^{2}} \right){{m}^{2}}-2hkm+{{k}^{2}}-{{b}^{2}}=0......\left( 1 \right) \\\ \end{aligned}$$ Equation 1 became a quadratic equation and hence we can use the property of quadratic equation through which we will get: $$\begin{aligned} & {{m}_{1}}{{m}_{2}}=\dfrac{c}{a} \\\ & \Rightarrow {{m}_{1}}{{m}_{2}}=\dfrac{{{k}^{2}}-{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}}........\left( 2 \right) \\\ \end{aligned}$$ $$\begin{aligned} & {{m}_{1}}+{{m}_{2}}=-\dfrac{b}{a} \\\ & \Rightarrow {{m}_{1}}+{{m}_{2}}=\dfrac{2hk}{{{h}^{2}}-{{a}^{2}}}........\left( 3 \right) \\\ \end{aligned}$$ We also know that we can write slope as $$\tan \left( \theta \right)$$ so we can also say that: $${{m}_{1}}=\tan \left( {{\theta }_{1}} \right)$$ and $${{m}_{2}}=\tan \left( {{\theta }_{2}} \right)$$ Now when we substitute values of $${{m}_{1}},{{m}_{2}}$$in equation 3 we will get: $$\begin{aligned} & \tan \left( {{\theta }_{1}} \right)+\tan \left( {{\theta }_{2}} \right)=\dfrac{2hk}{{{h}^{2}}-{{a}^{2}}}\left[ \text{on squaring both sides} \right] \\\ & \Rightarrow {{\left( \tan \left( {{\theta }_{1}} \right)+\tan \left( {{\theta }_{2}} \right) \right)}^{2}}={{\left( \dfrac{2hk}{{{h}^{2}}-{{a}^{2}}} \right)}^{2}}...........\left( 4 \right) \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow {{\left( \dfrac{2hk}{{{h}^{2}}-{{a}^{2}}} \right)}^{2}}=\lambda +2\dfrac{{{k}^{2}}-{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}} \\\ & \Rightarrow \left( \dfrac{4{{h}^{2}}{{k}^{2}}}{{{h}^{2}}-{{a}^{2}}} \right)=\lambda \left( {{h}^{2}}-{{a}^{2}} \right)+2\left( {{k}^{2}}-{{b}^{2}} \right) \\\ & \Rightarrow 4{{h}^{2}}{{k}^{2}}=\lambda {{\left( {{h}^{2}}-{{a}^{2}} \right)}^{2}}+2\left( {{k}^{2}}-{{b}^{2}} \right)\left( {{h}^{2}}-{{a}^{2}} \right) \\\ & \Rightarrow \lambda {{\left( {{x}^{2}}-{{a}^{2}} \right)}^{2}}=2\left( {{x}^{2}}{{y}^{2}}+{{y}^{2}}{{a}^{2}}+{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{b}^{2}} \right) \\\ \end{aligned}$$ Now, $$\Rightarrow {{\left( \tan \left( {{\theta }_{1}} \right)+\tan \left( {{\theta }_{2}} \right) \right)}^{2}}={{\tan }^{2}}\left( {{\theta }_{1}} \right)+{{\tan }^{2}}\left( {{\theta }_{2}} \right)+2\tan \left( {{\theta }_{1}} \right)\tan \left( {{\theta }_{2}} \right)..........\left( 5 \right)$$ And we know that $${{\tan }^{2}}\theta 1+{{\tan }^{2}}\theta 2=\lambda \left( \text{constant} \right)$$ that has been mentioned in the question, we will substitute the variable in the equation 4 and we get: $$\begin{aligned} & \Rightarrow {{\left( \tan \left( {{\theta }_{1}} \right)+\tan \left( {{\theta }_{2}} \right) \right)}^{2}}=\lambda +2\tan \left( {{\theta }_{1}} \right)\tan \left( {{\theta }_{2}} \right) \\\ & \Rightarrow {{\left( \tan \left( {{\theta }_{1}} \right)+\tan \left( {{\theta }_{2}} \right) \right)}^{2}}=\lambda +2{{m}_{1}}{{m}_{2}}.........\left( 6 \right) \\\ \end{aligned}$$ Now we will substitute the value of $${{m}_{1}}{{m}_{2}}$$ from equation 2 in equation 6 and we get, $$\Rightarrow {{\left( \tan \left( {{\theta }_{1}} \right)+\tan \left( {{\theta }_{2}} \right) \right)}^{2}}=\lambda +2\dfrac{{{k}^{2}}-{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}}$$ Now using equation 4 we will get: $$\begin{aligned} & \Rightarrow {{\left( \dfrac{2hk}{{{h}^{2}}-{{a}^{2}}} \right)}^{2}}=\lambda +2\dfrac{{{k}^{2}}-{{b}^{2}}}{{{h}^{2}}-{{a}^{2}}} \\\ & \Rightarrow \left( \dfrac{4{{h}^{2}}{{k}^{2}}}{{{h}^{2}}-{{a}^{2}}} \right)=\lambda \left( {{h}^{2}}-{{a}^{2}} \right)+2\left( {{k}^{2}}-{{b}^{2}} \right) \\\ & \Rightarrow 4{{h}^{2}}{{k}^{2}}=\lambda {{\left( {{h}^{2}}-{{a}^{2}} \right)}^{2}}+2\left( {{k}^{2}}-{{b}^{2}} \right)\left( {{h}^{2}}-{{a}^{2}} \right) \\\ & \Rightarrow \lambda {{\left( {{h}^{2}}-{{a}^{2}} \right)}^{2}}=2\left( {{h}^{2}}{{k}^{2}}+{{k}^{2}}{{a}^{2}}+{{b}^{2}}{{h}^{2}}-{{a}^{2}}{{b}^{2}} \right) \\\ \end{aligned}$$ Now we will substitute the value of h and k with x and y to give the required answer i.e. locus of point P which is: $$\lambda {{\left( {{x}^{2}}-{{a}^{2}} \right)}^{2}}=2\left( {{x}^{2}}{{y}^{2}}+{{y}^{2}}{{a}^{2}}+{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{b}^{2}} \right)$$ The locus of point P is $$\lambda {{\left( {{x}^{2}}-{{a}^{2}} \right)}^{2}}=2\left( {{x}^{2}}{{y}^{2}}+{{y}^{2}}{{a}^{2}}+{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{b}^{2}} \right)$$ **Note:** In the type of question that is mentioned above the main problem that arises is at the point on where to use the relation i.e. mentioned in the question, as in the above question the relation that was stated was between slopes and not any trigonometric equation, so try to remember what does each trigonometric relation states and what does they signify.