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Hyperbola question

2023 · 6 Apr · Shift 2 · Q38
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  5. /2023 · 6 Apr · Shift 2 · Q38

Hyperbola question

2023 · 6 Apr · Shift 2 · Q38

JEE MainMathematicsHyperbolaNumerical+4 / −1
Let the eccentricity of an ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1a2x2​+b2y2​=1 is reciprocal to that of the hyperbola 2x2−2y2=12 x^{2}-2 y^{2}=12x2−2y2=1. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. Given hyperbola and its eccentricity

The hyperbola is 2x2−2y2=1  ⟹  x21/2−y21/2=1.2x^2-2y^2=1 \implies \frac{x^2}{1/2}-\frac{y^2}{1/2}=1.2x2−2y2=1⟹1/2x2​−1/2y2​=1.

So for this hyperbola, ah2=12,bh2=12.a_h^2=\frac12, \qquad b_h^2=\frac12.ah2​=21​,bh2​=21​.

Its eccentricity is eh=1+bh2ah2=1+1=2.e_h=\sqrt{1+\frac{b_h^2}{a_h^2}}=\sqrt{1+1}=\sqrt{2}.eh​=1+ah2​bh2​​​=1+1​=2​.

  1. Eccentricity of the ellipse

The ellipse has eccentricity reciprocal to that of the hyperbola, so e=12.e=\frac{1}{\sqrt{2}}.e=2​1​.

For the ellipse x2a2+y2b2=1,a>b,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b,a2x2​+b2y2​=1,a>b, we know e2=1−b2a2.e^2=1-\frac{b^2}{a^2}.e2=1−a2b2​.

Thus, 12=1−b2a2  ⟹  b2a2=12.\frac12=1-\frac{b^2}{a^2} \implies \frac{b^2}{a^2}=\frac12.21​=1−a2b2​⟹a2b2​=21​. So, b2=a22.b^2=\frac{a^2}{2}.b2=2a2​.

  1. Condition for orthogonal intersection

Let F(x,y)=x2a2+y2b2−1=0F(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0F(x,y)=a2x2​+b2y2​−1=0 and G(x,y)=2x2−2y2−1=0.G(x,y)=2x^2-2y^2-1=0.G(x,y)=2x2−2y2−1=0.

Two curves intersect orthogonally if at a point of intersection, ∇F⋅∇G=0.\nabla F \cdot \nabla G=0.∇F⋅∇G=0.

Now,

\qquad \nabla G=(4x,-4y).$$ Hence, $$\nabla F\cdot \nabla G=\frac{8x^2}{a^2}-\frac{8y^2}{b^2}=0$$ which gives $$\frac{x^2}{a^2}=\frac{y^2}{b^2}.$$ Since the point also lies on the ellipse, $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$ so each term must be $\frac12$: $$\frac{x^2}{a^2}=\frac12, \qquad \frac{y^2}{b^2}=\frac12.$$ Thus, $$x^2=\frac{a^2}{2}, \qquad y^2=\frac{b^2}{2}.$$ 4. **Use the hyperbola equation** Substitute into $$2x^2-2y^2=1:$$ $$2\left(\frac{a^2}{2}\right)-2\left(\frac{b^2}{2}\right)=1$$ $$a^2-b^2=1.$$ But from step 2, $$b^2=\frac{a^2}{2}.$$ So, $$a^2-\frac{a^2}{2}=1 \implies \frac{a^2}{2}=1 \implies a^2=2.$$ Hence, $$b^2=1.$$ 5. **Length of latus rectum of the ellipse** For the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$ the length of the latus rectum is $$L=\frac{2b^2}{a}.$$ Therefore, $$L^2=\left(\frac{2b^2}{a}\right)^2.

Using a2=2a^2=2a2=2 and b2=1b^2=1b2=1, L2=4⋅12=2.L^2=\frac{4\cdot 1}{2}=2.L2=24⋅1​=2.

So the required square of the length of the latus rectum is 2.\boxed{2}. 2​.

  1. Comparison with stored answer

Derived answer is 222, which matches the stored correct answer.

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