JEE Mathematics 2023 Question with Solution

Given: The curves are $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$, and their points of intersection lie on $y = 1$. Also, $a,b,c,d,e,f$ are positive and in G.P.

Find: Which statement among the options is correct.

From the first parabola, putting $y=1$ gives

$$ax^2 + 2bx + c = 0$$

Since $a,b,c$ are in G.P.,

$$b^2 = ac$$

Therefore,

$$ax^2 + 2\sqrt{ac}\,x + c = 0$$

which can be written as

$$\left(x\sqrt{a}+\sqrt{c}\right)^2 = 0$$

Hence the common intersection point on $y=1$ has

$$x^2 = \frac{a}{c}, \qquad x = -\sqrt{\frac{c}{a}}$$

Using this in the second parabola with $y=1$,

$$dx^2 + 2ex + f = 0$$

so

$$d\left(\frac{c}{a}\right) + 2e\left(-\sqrt{\frac{c}{a}}\right) + f = 0$$

Therefore,

$$\frac{dc}{a} + f = 2e\sqrt{\frac{c}{a}}$$

Dividing by $cf$,

$$\frac{d}{a} + \frac{f}{c} = 2\frac{e}{\sqrt{ac}}$$

Now, since $a,b,c,d,e,f$ are in G.P., we have

$$b^2 = ac \quad \Rightarrow \quad b = \sqrt{ac}$$

Thus,

$$\frac{d}{a} + \frac{f}{c} = 2\frac{e}{b}$$

So $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

Therefore, the correct option is C.

The solution labels the option as D, but the written concluding statement matches option C in the given options.

Given: On the line $y=1$, the first parabola becomes a quadratic in $x$ with coefficients $a,b,c$ in G.P.

Find: The relation among $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$.

Because $a,b,c$ are in G.P.,

$$b^2 = ac$$

So the quadratic

$$ax^2 + 2bx + c = 0$$

has discriminant

$$(2b)^2 - 4ac = 0$$

Hence it has a repeated root. Since the two parabolas intersect on $y=1$, that repeated root must also satisfy

$$dx^2 + 2ex + f = 0$$

Substituting the repeated root relation leads to

$$\frac{d}{a} + \frac{f}{c} = 2\frac{e}{b}$$

which is exactly the condition that

$$\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$$

are in A.P.

Therefore, the correct option is C.