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QJEEAdv18Mac5a6: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Mac5a6 Solution: Consider two straight lines, each of which is tangent to both…

Consider two straight lines, each of which is tangent to both the circle x2 + y2 = (1/2) and the parabola y2 = 4x. Let these lines intersect at the point Q. Consider the ellipse whose centre is at the origin O(0, 0) and whose semi-major axis is OQ. If the length of the minor axis of this ellipse is 2\sqrt 2 , then which of the following statement(s) is (are) TRUE?

Multiple Choice Options

  • A For the ellipse, the eccentricity is 1 2\sqrt 2 and the length of the latus rectum is 1
  • B For the ellipse, the eccentricity is 1/2 and the length of the latus rectum is 1/2
  • C The area of the region bounded by the ellipse between the lines x=12x = {1 \over {\sqrt 2 }} and x = 1 is 142(π−2){1 \over {4\sqrt 2 }}(\pi - 2)
  • D The area of the region bounded by the ellipse between the lines x=12x = {1 \over {\sqrt 2 }} and x = 1 is 116(π−2){1 \over {16}}(\pi - 2)

Step-by-Step Verified Solution

We have,

Equations of circle

x2+y2=12{x^2} + {y^2} = {1 \over 2}

and Equations of parabola

y2 = 4x

JEE Advanced 2018 Paper 2 Offline Mathematics - Ellipse Question 6 English Explanation 1

Let the equation of common tangent of parabola and circle is

y=mx+1my = mx + {1 \over m}

Since, radius of circle = 12{1 \over {\sqrt 2 }}

∴\therefore 12{1 \over {\sqrt 2 }} = ∣0+0+1m1+m2∣\left| {{{0 + 0 + {1 \over m}} \over {\sqrt {1 + {m^2}} }}} \right|

⇒\Rightarrow m4 + m2 −- 2 = 0

⇒\Rightarrow m = ±\pm 1

∴\therefore Equation of common tangents are

y = x + 1 and y = −- x −- 1

Intersection point of common tangent at Q( −- 1, 0)

∴\therefore Equation of ellipse

x21+y21/2=1{{{x^2}} \over 1} + {{{y^2}} \over {1/2}} = 1

where, a2=1{a^2} = 1 , b2=1/2{b^2} = 1/2

Now, eccentricity

=1−b2a2=1−12=12= \sqrt {1 - {{{b^2}} \over {{a^2}}}} = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}

and length of latusrectum

=2b2a=2(12)1=1= {{2{b^2}} \over a} = {{2\left( {{1 \over 2}} \right)} \over 1} = 1

JEE Advanced 2018 Paper 2 Offline Mathematics - Ellipse Question 6 English Explanation 2

∴\therefore Area of shaded region

=2∫1/21121−x2dx= 2\int_{1/\sqrt 2 }^1 {{1 \over {\sqrt 2 }}} \sqrt {1 - {x^2}} dx

=2[x21−x2+12sin−1x]1/21=2[(0+π4)−(14+π8)]

=2[(0+π4)−(14+π8)]= \sqrt 2 \left[ {\left( {0 + {\pi \over 4}} \right) - \left( {{1 \over 4} + {\pi \over 8}} \right)} \right]

=2(π8−14)=π−242= \sqrt 2 \left( {{\pi \over 8} - {1 \over 4}} \right) = {{\pi - 2} \over {4\sqrt 2 }}

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