E:$$x2a2+y2b2=1$$ and P:$$y2=4bx$$,ab. Statement $$1$$: The tangent at the positive end of the minor axis of the ellipse. E passes through the positive end of the latus rectum of the parabola P. Statement $$2$$: If the latus rectum of the parabola P is same as that of the ellipse E, then eccentricity of E is $$1/2$$

Question

E:$$x2a2+y2b2=1$$ and  P:$$y2=4bx$$,a>b. Statement $$1$$: The tangent at the positive end of the minor axis of the ellipse. E passes through the positive end of the latus rectum of the parabola P.  Statement $$2$$: If the latus rectum of the parabola P is same as that of the ellipse  E, then eccentricity of  E  is  $$1/2$$

Infinity Learn · Accepted Answer

Tangent at the positive end of the minor axis of  E is y $$=$$ b which meets the parabola  $$y2$$ $$=$$ $$4bx$$  at the $$pointb4$$,b   which is not an end of the latus rectum of  P.  so $$statement-$$ I is false. In $$statement-2$$  if  x $$=$$ b  and   x $$=$$ ae represent the same line then   b $$=$$ ae ⇒$$ba=e$$⇒$$a21$$−$$e2=a2e2$$⇒$$e2=12$$⇒$$e=12$$.Thus $$statement-2$$ is true.

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E:x2a2+y2b2=1 and P:y2=4bx,a>b. Statement 1: The tangent at the positive end of the minor axis of the ellipse. E passes through the positive end of the latus rectum of the parabola P. Statement 2: If the latus rectum of the parabola P is same as that of the ellipse E, then eccentricity of E is 1/2

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