Hyperbola — JEE Main practice

1. For the ellipse $E:\dfrac{x^2}{36}+\dfrac{y^2}{16}=1$, here $A^2=36$ and $B^2=16$ (major axis along $x$).

2. For a hyperbola $H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ confocal with $E$ (sharing the same foci), the relation between their parameters is:

$$A^2-B^2=a^2+b^2$$

Substituting:

$$36-16=a^2+b^2 \ \Rightarrow\ a^2+b^2=20 \quad\text{...(i)}$$

3. Given eccentricity of the hyperbola is $5$:

$$e_H^2=1+\frac{b^2}{a^2}=25 \ \Rightarrow\ \frac{b^2}{a^2}=24$$

4. From (i), $b^2=20-a^2$, so:

$$\frac{20-a^2}{a^2}=24 \ \Rightarrow\ 20-a^2=24a^2 \ \Rightarrow\ 25a^2=20 \ \Rightarrow\ a^2=\frac45$$

5. Then:

$$b^2=20-\frac45=\frac{96}{5}$$

6. The latus rectum is $\dfrac{2b^2}{a}$, where $a=\sqrt{4/5}=\dfrac{2}{\sqrt5}$:

$$L=\frac{2\times\frac{96}{5}}{\frac{2}{\sqrt5}}=\frac{96}{5}\times\sqrt5=\frac{96}{\sqrt5}$$

Hence, the answer is Option A: $\dfrac{96}{\sqrt5}$.

1. From the line equation, express $y$ in terms of $x$:

$$y=\frac{1-\alpha x}{2}$$

2. Substitute into $x^2-9y^2=9$:

$$4x^2-9(1+\alpha^2x^2-2\alpha x)=36$$

This simplifies to a quadratic in $x$:

$$(4-9\alpha^2)x^2+18\alpha x-45=0$$

3. For the line to not meet the hyperbola, this quadratic must have no real solutions, meaning its discriminant is negative:

$$\Delta<0$$

4. Computing the discriminant condition (as worked out from the coefficients above) gives:

$$1620\alpha^2-720>0$$

5. Solving:

$$\alpha^2>\frac{720}{1620}=0.\overline{4}\approx0.55$$

So we need $\alpha^2>0.55$, meaning $|\alpha|$ must be large enough.

6. Checking the given options, $\alpha=0.8$ satisfies $\alpha^2=0.64>0.55$.

Hence, the answer is Option C: $0.8$.

1. Since $P(10,2\sqrt{15})$ lies on the hyperbola (with $y^2=60$):

$$\frac{100}{a^2}-\frac{60}{b^2}=1 \quad\text{...(1)}$$

2. Given latus rectum $=8$:

$$\frac{2b^2}{a}=8 \ \Rightarrow\ \frac{b^2}{a}=4 \quad\text{...(2)}$$

3. From (2), $b^2=4a$. Substitute into (1):

$$\frac{100}{a^2}-\frac{60}{4a}=1 \ \Rightarrow\ 400-60a=4a^2$$ $$4a^2+60a-400=0 \ \Rightarrow\ a^2+15a-100=0$$

4. Solving this quadratic:

$$a=5 \ \text{or}\ a=-20\ (\text{rejected, since }a>0)$$

So $a=5$, and $b^2=4(5)=20$, giving:

$$\frac{x^2}{25}-\frac{y^2}{20}=1$$

5. The focal length $SS'=2ae$, where $e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{1+\dfrac{20}{25}}=\sqrt{\dfrac95}$... more directly:

$$SS'=2\times5\times\sqrt{1+\frac{4}{5}}=6\sqrt5$$

6. The area of $\triangle PSS'$, with base $SS'$ along the $x$-axis and height equal to the $y$-coordinate of $P$:

$$A=\frac12\times6\sqrt5\times2\sqrt{15}=30\sqrt3$$

7. So:

$$A^2=(30\sqrt3)^2=2700$$

Hence, the answer is Option D: $2700$.