JEE Advance - Mathematics (2020 - Paper 1 Offline - No. 4)
Let a, b and $$\lambda $$ be positive real numbers. Suppose P is an end point of the latus return of the
parabola y2 = 4$$\lambda $$x, and suppose the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ passes through the point P. If the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is
parabola y2 = 4$$\lambda $$x, and suppose the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ passes through the point P. If the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is
$${1 \over {\sqrt 2 }}$$
$${{1 \over 2}}$$
$${{1 \over 3}}$$
$${{2 \over 5}}$$
Explanation
Equation of given parabola is
y2 = 4$$\lambda $$x ....(i)
So the end point of the latus rectum of the parabola (i), P($$\lambda $$, 2$$\lambda $$) and the given ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, passes through point P($$\lambda $$, 2$$\lambda $$).
On differentiating the equation of parabola, w.r.t. 'x', we get
$${{dy} \over {dx}} = {{2\lambda } \over y}$$
$$ \therefore $$ Slope of tangent to the parabola at point P is m1 = 1
Similarly, on differentiating the equation of given ellipse, $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, w.r.t.x, we get $${{2x} \over {{a^2}}} + {{2y} \over {{b^2}}}{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = - {{{b^2}x} \over {{a^2}y}}$$
$$ \therefore $$ Slope of tangent to the ellipse at point P is $${m_2} = - {{{b^2}} \over {2{a^2}}}$$
$$ \because $$ It is given that the tangents are perpendicular to each other. So, $${m_1}{m_2} = - 1$$
$$ \Rightarrow (1)\left( { - {{{b^2}} \over {2{a^2}}}} \right) = - 1$$
$$ \Rightarrow {{{b^2}} \over {{a^2}}} = 2 \Rightarrow b = \sqrt 2 a$$
$$ \therefore $$ Eccentricity of ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ will be
$$e = \sqrt {1 - {{{a^2}} \over {{b^2}}}} = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}$$ {$$ \because $$ b > a}
y2 = 4$$\lambda $$x ....(i)
So the end point of the latus rectum of the parabola (i), P($$\lambda $$, 2$$\lambda $$) and the given ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, passes through point P($$\lambda $$, 2$$\lambda $$).
On differentiating the equation of parabola, w.r.t. 'x', we get
$${{dy} \over {dx}} = {{2\lambda } \over y}$$
$$ \therefore $$ Slope of tangent to the parabola at point P is m1 = 1
Similarly, on differentiating the equation of given ellipse, $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, w.r.t.x, we get $${{2x} \over {{a^2}}} + {{2y} \over {{b^2}}}{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = - {{{b^2}x} \over {{a^2}y}}$$
$$ \therefore $$ Slope of tangent to the ellipse at point P is $${m_2} = - {{{b^2}} \over {2{a^2}}}$$
$$ \because $$ It is given that the tangents are perpendicular to each other. So, $${m_1}{m_2} = - 1$$
$$ \Rightarrow (1)\left( { - {{{b^2}} \over {2{a^2}}}} \right) = - 1$$
$$ \Rightarrow {{{b^2}} \over {{a^2}}} = 2 \Rightarrow b = \sqrt 2 a$$
$$ \therefore $$ Eccentricity of ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ will be
$$e = \sqrt {1 - {{{a^2}} \over {{b^2}}}} = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}$$ {$$ \because $$ b > a}
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