Question:
Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of π/2 at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse \(E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a^2>b^2.\) If e is the eccentricity of the ellipse E, then the value of \(\frac{1}{e^2}\) is equal to
Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of π/2 at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse \(E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a^2>b^2.\) If e is the eccentricity of the ellipse E, then the value of \(\frac{1}{e^2}\) is equal to
Updated On: Mar 4, 2024
- \(1+\sqrt2\)
- \(3+2\sqrt2\)
- \(1+2\sqrt3\)
- \(4+5\sqrt3\)
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The Correct Option is B
Solution and Explanation
The correct option is(B): \(3+2\sqrt2\)
\(∠PQR=\frac{\pi}{2}\)
∴ P ≡ (1, 2) & Q(1, –2)
∴ for ellipse
\(\frac{1}{a^2}+\frac{4}{b^2}=1\)
and ae = 1
⇒ (5 – e2)e2 = 1 – e2
⇒ e4 – 6e2 + 1 = 0
\(⇒\frac{1}{e^2}=3+2\sqrt2\)
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.