Question

If tangents are drawn to the ellipse x² + 2y² = 2, then the locus of the mid-point of the intercept made by the tangents between the co-ordinate axes is –

• A. $\frac{1}{2x^{2}} + \frac{1}{4y^{2}}$ = 1
• B. $\frac{1}{4x^{2}} + \frac{1}{2y^{2}} = 1$
• C. $\frac{x^{2}}{2} + \frac{y^{2}}{4} = 1$
• D. $\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1$

Answer 1

Answer: (A)

$\frac{1}{2x^{2}} + \frac{1}{4y^{2}}$ = 1

Answer 2

$\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$ $\left\lbrack \begin{matrix} a = \sqrt{2} \\ b = 1 \end{matrix} \right.\$

Eq. of tangent at point R ̃ T = 0

$\left[ \begin{array} { l } \frac { x ( \sqrt { 2 } \cos \phi ) } { 2 } + \frac { y ( 1 \sin \phi ) } { 1 } = 1 \\ x = 0 \Rightarrow y = \frac { 1 } { \sin \phi } \\ y = 0 \Rightarrow x = \frac { \sqrt { 2 } } { \cos \phi } \end{array} \right.$

Locus of mid point of PQ ̃ PQ ̃ PM = QM

M (h, k) ̃ Locus = ?

M$\left\{ \begin{matrix} h = \left( \frac{\sqrt{2}}{\cos\varphi} + 0 \right)/2 \Rightarrow \cos\varphi = \frac{1}{\sqrt{2}h} \\ k = \left( \frac{1}{\sin\varphi} + 0 \right)/2 \Rightarrow \sin\varphi = \frac{1}{2k} \end{matrix} \right.\$

̃ cos² f + sin² f = 1 ̃ $\frac{1}{2x^{2}} + \frac{1}{4y^{2}}$ = 1