Parabola — JEE Main practice

1. Let $O(0,0)$ be the vertex and $Q$ a point on $x^2=4y$, parametrized as $Q(2t,t^2)$.

2. Let $P(x,y)$ divide $OQ$ internally in the ratio $2:3$. Using the section formula:

$$x=\frac{2(2t)+3(0)}{5}=\frac{4t}{5},\qquad y=\frac{2(t^2)+3(0)}{5}=\frac{2t^2}{5}$$

3. Eliminate the parameter $t$. From the first equation, $t=\frac{5x}{4}$. Substituting into the second:

$$y=\frac25\left(\frac{5x}{4}\right)^2=\frac25\cdot\frac{25x^2}{16}=\frac{5x^2}{8}$$

4. So the conic C is:

$$5x^2=8y$$

5. The equation of a chord of a conic $S=0$ bisected at a point $(x_1,y_1)$ follows the standard "T = S1" rule. Here $S\equiv5x^2-8y$, and $(x_1,y_1)=(1,2)$:

$$S_1=5(1)^2-8(2)=5-16=-11$$

6. Construct $T$ by replacing $x^2$ with $xx_1$ and $y$ with $\frac{y+y_1}{2}$:

$$T=5x(1)-8\left(\frac{y+2}{2}\right)=5x-4(y+2)=5x-4y-8$$

7. Setting $T=S_1$:

$$5x-4y-8=-11 \;\Rightarrow\; 5x-4y+3=0$$

Hence, the answer is Option A.

1. For $y^2=16x$, $4a=16 \Rightarrow a=4$. Using parametric coordinates $(4t^2,8t)$, the point $(16,16)$ corresponds to $t_1$ where $4t_1^2=16$ and $8t_1=16$, giving $t_1=2$.

2. For a focal chord, the two endpoints satisfy $t_1t_2=-1$:

$$2t_2=-1 \;\Rightarrow\; t_2=-\frac12$$

3. So the other endpoint is:

$$B=\left(4\left(\frac14\right),8\left(-\frac12\right)\right)=(1,-4)$$

4. Since $P(\alpha,\beta)$ divides the focal chord (with endpoints $(16,16)$ and $(1,-4)$) internally in the ratio $5:2$, there are two possible orderings, giving two possible points via the section formula:

$$P=\left(\frac{37}{7},\frac{12}{7}\right)\ \text{or}\ \left(\frac{82}{7},\frac{72}{7}\right)$$

5. Computing $\alpha+\beta$ for each:

$$\frac{37}{7}+\frac{12}{7}=7,\qquad \frac{82}{7}+\frac{72}{7}=22$$

6. The minimum of these two values is:

$$7$$

Hence, the answer is Option D.

1. For $y^2=12x$, $4a=12 \Rightarrow a=3$, so let $P=(3t^2,6t)$ be a point on the parabola using parametric coordinates.

2. Let $A=(x,0)$ lie on the $x$-axis. Since $\angle OPA=90°$, the product of the slopes of $OP$ and $PA$ is $-1$:

$$\frac{6t}{3t^2}\times\frac{6t}{3t^2-x}=-1$$

3. Simplifying:

$$\frac{2}{t}\times\frac{6t}{3t^2-x}=-1 \;\Rightarrow\; 12=x-3t^2$$

So $x=12+3t^2$.

4. Now find the centroid of $\triangle OPA$ with $O(0,0)$, $P(3t^2,6t)$, $A(x,0)$:

$$h=\frac{x+3t^2+0}{3},\qquad k=\frac{0+0+6t}{3}=2t$$

5. Substituting $x=12+3t^2$:

$$3h=12+6t^2 \;\Rightarrow\; h=4+2t^2$$

6. Since $k=2t \Rightarrow t=\frac{k}{2}$, substitute into $h=4+2t^2$:

$$h=4+2\left(\frac{k}{2}\right)^2=4+\frac{k^2}{2}$$

7. Replacing $h$ by $x$ and $k$ by $y$:

$$x=4+\frac{y^2}{2} \;\Rightarrow\; 2x=8+y^2 \;\Rightarrow\; y^2-2x+8=0$$

Hence, the answer is Option C.