Question

The tangents drawn from a point P to the ellipse makes an angle $${\theta }_1$$and $${\theta }_2$$with the major axis, find the locus of P when
$$\tan {\theta }_1+\tan {\theta }_2$$is constant$$=c$$

Answer 1

Hint: Use the standard equation of tangent for ellipse $$y=mx\pm \sqrt{a^2{m}^2+b^2}$$, then form a quadratic in m and proceed further and use given information.

Complete step-by-step answer:

Let given ellipse $$={\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1(\because a>b)$$

Hence, the X-axis is the major axis.

Equation (2) will pass through (h, k): -

$k=mh\pm \sqrt{a^2{m}^2+b^2}$

${(k-mh)}^2=a^2{m}^2+b^2$

$k^2+m^2{h}^2-2mkh=a^2{m}^2+b^2$

$m^2(h^2-a^2)-2mkh+k^2-b^2=0-(3)$

As m is quadratic function, hence it have two roots or two tangents passing through P are there with slopes $$\tan {\theta }_1$$and $$\tan {\theta }_2$$.

Hence, $$\tan {\theta }_1$$and $$\tan {\theta }_2$$are roots of quadratic obtained.

$$\tan {\theta }_1+\tan {\theta }_2={\displaystyle \frac{2hk}{h^2-a^2}}-(4)$${sum of roots$$={\displaystyle \frac{-b}{a}}$$in $$a{x}^2+bx+c=0$$}

$$\tan {\theta }_1.\tan {\theta }_2={\displaystyle \frac{k^2-b^2}{h^2-a^2}}-(5)$${Product of roots is $${\displaystyle \frac{c}{a}}$$ in $$a{x}^2+bx+c=0$$}

Replacing (h, k) by (x, y) to get locus: -

$${\displaystyle \frac{2xy}{x^2-a^2}}=c$$ {Replacing $$\tan {\theta }_1+\tan {\theta }_2=c$$as it is given}

$$(x^2-a^2)c=2xy$$

$$c{x}^2-2xy-c{a}^2=0$$ is required locus.

Note: Need to observe the relation from quadratic formed to eliminate $${\theta }_1\mathrm{\&}{\theta }_2$$by using the given condition $$\tan {\theta }_1+\tan {\theta }_2=c$$is key point of the equation which is done by sum of roots of quadratic found after solving $$y=mx\pm \sqrt{a^2{m}^2+b^2}$$.

Direct using the formula of tangent to an ellipse whose slope is given, solved the problem easily. Using formulae in conic sections always helps and makes the solution easier. Here, we can use direct formula of tangent of ellipse i.e. $$y=mx\pm \sqrt{a^2{m}^2+b^2}$$standard equation $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$$which can be proved by following approach: -

Now, y=mx + c and $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$$have only one intersection point (touching the ellipse). So, if we substitute y=mx + c in $${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$$at place of y then we will get quadratic in x which should have one solution (as tangent and ellipse have only one intersection point). Then we will get $$c=\pm \sqrt{a^2{m}^2+b^2}$$by making decrement of that quadratic to 0.