Ellipse Ques 24
- Find the equation of the common tangent in 1st quadrant to the circle $x^{2}+y^{2}=16$ and the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{4}=1$. Also, find the length of the intercept of the tangent between the coordinate axes.
(2005, 4M)
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Answer:
Correct Answer: 24.$y= - \frac {2x}{\sqrt 3} + 4 \sqrt \frac {7}{3}, \frac {14 \sqrt 3}{3} $
Solution:
- Let the common tangent to $x^{2}+y^{2}=16$ and $\frac{x^{2}}{25}+\frac{y^{2}}{4}=1$ be
$ \begin{alignedat} & y=m x+4 \sqrt{1+m^{2}} \\ & \text { and } \quad y=m x+\sqrt{25 m^{2}+4} \end{aligned} $
Since, Eqs. (i) and (ii) are the same tangent.
$ \begin{alignedat} & \therefore \quad 4 \sqrt{1+m^{2}}=\sqrt{25 m^{2}+4} \\ & \Rightarrow \quad 16\left(1+m^{2}\right)=25 m^{2}+4 \\ & \Rightarrow \quad 9 m^{2}=12 \text{ m}^{2} \ & \Rightarrow \quad m= \pm 2 / \sqrt{3} \end{aligned} $
Since, tangent is in the first quadrant.
$ \begin{array}{ll} \therefore & m<0 \\ \Rightarrow & m=-2 / \sqrt{3} \end{array} $
So, the equation of the common tangent is
$ y=-\frac{2 x}{\sqrt{3}}+4 \sqrt{\frac{7}{3}} $
which meets coordinate axes at $A(2 \sqrt{7}, 0)$ and $B\left(0,4 \sqrt{\frac{7}{3}}\right)$.
$ \begin{alignedat} \therefore \quad A B & =\sqrt{(2 \sqrt{7}-0)^{2}+(0-4 \sqrt{\frac{7}{3}})^{2}} \\ & =\sqrt{28+\frac{11}{3}}=\sqrt{\frac{196}{3}}=\frac{14}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{14 \sqrt{3}}{3} \end{aligned} $
BETA
