Ellipse — JEE Main practice

1. Write the ellipse in standard form: $3x^2+4y^2=1$ means $\frac{x^2}{1/3}+\frac{y^2}{1/4}=1$, so $a^2=\frac13$, $b^2=\frac14$.

2. For the line $\alpha x+4y-\sqrt7=0$ to be tangent to this ellipse, use the tangency condition $c^2=a^2m^2+b^2$ after writing the line in slope form. Applying this condition here gives:

$$\frac{7}{16}=\frac13\cdot\frac{\alpha^2}{16}+\frac14 \;\Rightarrow\; \alpha=\pm3$$

3. Taking $\alpha=3$, the tangent line is $3x+4y-\sqrt7=0$. Working out the point of contact for a tangent line touching this ellipse gives:

$$P=\left(\frac{1}{\sqrt7},\frac{1}{\sqrt7}\right)$$

4. The eccentricity of the ellipse is:

$$e=\sqrt{1-\frac{1/4}{1/3}}=\sqrt{1-\frac34}=\frac12$$

5. The focal distance of a point from a focus equals $e$ times the distance from that point to the corresponding directrix — a defining property of conics. If $M$ and $M'$ are the feet of perpendiculars from $P$ to the two directrices at $x=\pm\frac{a}{e}$, then:

$$PS=e\left(\frac{a}{e}-\frac{1}{\sqrt7}\right),\qquad PS'=e\left(\frac{a}{e}+\frac{1}{\sqrt7}\right)$$

6. Substituting $a=\frac{1}{\sqrt3}$ and $e=\frac12$:

$$PS=\frac12\left(\frac{2}{\sqrt3}-\frac{1}{\sqrt7}\right)=\frac{1}{\sqrt3}-\frac{1}{2\sqrt7}$$ $$PS'=\frac12\left(\frac{2}{\sqrt3}+\frac{1}{\sqrt7}\right)=\frac{1}{\sqrt3}+\frac{1}{2\sqrt7}$$

Hence, one focal distance of $P$ is $\frac{1}{\sqrt3}+\frac{1}{2\sqrt7}$ — Option C.

1. Since $P$ lies on the ellipse, $\frac{\alpha^2}{25}+\frac{\beta^2}{9}=1$.

2. By the defining property of an ellipse, the sum of distances from any point to the two foci equals $2a$.

$$PS+PS'=2a=10$$

3. We are given $(PS)^2+(PS')^2-PS\cdot PS'=37$. Using the identity $(PS+PS')^2=(PS)^2+(PS')^2+2\,PS\cdot PS'$, rewrite the left side in terms of $PS+PS'$ and $PS\cdot PS'$.

$$(PS+PS')^2-3\,PS\cdot PS'=37$$

4. Substituting $PS+PS'=10$:

$$100-3\,PS\cdot PS'=37 \;\Rightarrow\; PS\cdot PS'=21$$

5. For an ellipse, the focal distances of a point $(\alpha,\beta)$ can be written as $a+e\alpha$ and $a-e\alpha$, where $a=5$ and $e=\frac{c}{a}=\frac45$ here (since $c^2=25-9=16$). So:

$$PS\cdot PS'=a^2-e^2\alpha^2=25-\frac{16}{25}\alpha^2=21$$

6. Solving:

$$\frac{16}{25}\alpha^2=4 \;\Rightarrow\; \alpha^2=\frac{25}{4}$$

7. Substituting back into the ellipse equation to find $\beta^2$:

$$\beta^2=9\left(1-\frac{\alpha^2}{25}\right)=9\left(1-\frac14\right)=\frac{27}{4}$$

8. Therefore,

$$\alpha^2+\beta^2=\frac{25}{4}+\frac{27}{4}=\frac{52}{4}=13$$

Hence, the answer is Option B.

1. Substitute $y=x+1$ into the ellipse equation $\frac{x^2}{2}+y^2=1$ to find the intersection points.

$$\frac{x^2}{2}+(x+1)^2=1$$

2. Expanding and simplifying gives a quadratic in $x$.

$$3x^2+4x=0$$

3. Solving gives $x=0$ or $x=-\frac43$, so the points are:

$$A(0,1),\quad B\left(-\frac43,-\frac13\right)$$

4. The angle $\angle AOB$ at the origin can be found from the directions of $OA$ and $OB$. The angle that $OB$ makes relative to the axis works out through its slope:

$$\theta=\tan^{-1}\left(\frac{-1/3}{-4/3}\right)=\tan^{-1}\left(\frac14\right)$$

5. Since $OA$ lies along the $y$-axis and $OB$ makes angle $\theta$ measured from the negative $x$-axis direction, the full angle between them is:

$$\angle AOB=\frac{\pi}{2}+\tan^{-1}\left(\frac14\right)$$

Hence, the answer is Option A.