If the normal at the end of latus rectum of an ellipse x2-a2-y2-b2-1 of eccen-tricity e passes through one end of the minor axis then e4-e2-A- 0B- 1C- -1D- 2

The correct option is B Latus rectum = (ae, b^2/a ) Equation of normal is a^2x/ae b^2y/b^2/a = a^2 b^2 ⇒ax/e ay = a^2 b^2 Since it passes through (0, ...

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Standard XI

Mathematics

Common Ratio

If the normal...

Question

If the normal at the end of latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ of eccen-tricity e passes through one end of the minor axis then $e^{4} + e^{2} =$

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Solution

The correct option is B
Latus rectum $= (a e, \frac{b^{2}}{a})$
Equation of normal is $\frac{a^{2} x}{a e} - \frac{b^{2} y}{\frac{b^{2}}{a}} = a^{2} - b^{2} \Rightarrow \frac{a x}{e} - a y = a^{2} - b^{2}$
Since it passes through (0, b) then $- a b = a^{2} - b^{2} \Rightarrow \sqrt{1 - e^{2}} = e^{2} \Rightarrow 1 - e^{2} = e^{4} \Rightarrow e^{2} = 1$

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If the normal at the end of latus rectum of an ellipse x2a2+y2b2=1 of eccen-tricity e passes through one end of the minor axis then e4+e2=

The correct option is B Latus rectum =(ae,b2a) Equation of normal is a2xae−b2yb2a=a2−b2⇒axe−ay=a2−b2 Since it passes through (0, b) then −ab=a2−b2⇒√1−e2=e2⇒1−e2=e4⇒e2=1

If the normal at the end of latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ of eccen-tricity e passes through one end of the minor axis then $e^{4} + e^{2} =$