JEE Mathematics 2025 Question with Solution

Given: The ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and the midpoint of the chord is $(3,1)$.

Find: The equation of the chord.

For the ellipse $S \equiv \frac{x^2}{25} + \frac{y^2}{16} - 1 = 0$, the equation of the chord whose midpoint is $(x_1,y_1)$ is obtained by using the midpoint form:

$$T = S_1$$

where

$$T = \frac{xx_1}{25} + \frac{yy_1}{16} - 1$$

and

$$S_1 = \frac{x_1^2}{25} + \frac{y_1^2}{16} - 1$$

Now substitute $x_1 = 3$ and $y_1 = 1$:

$$\frac{3x}{25} + \frac{y}{16} - 1 = \frac{9}{25} + \frac{1}{16} - 1$$ $$\frac{3x}{25} + \frac{y}{16} = \frac{9}{25} + \frac{1}{16}$$ $$\frac{3x}{25} + \frac{y}{16} = \frac{144 + 25}{400} = \frac{169}{400}$$

Multiplying by $400$,

$$48x + 25y = 169$$

Therefore, the equation of the chord is $48x + 25y = 169$. So the defensible correct option from the given choices is D. The solution also contains a conflicting final selection of C, but the midpoint formula working gives D.

Given: The midpoint is $(3,1)$.

Find: Which option can represent a chord with this midpoint.

A chord with midpoint $(3,1)$ must have its midpoint lying on the line itself. So substitute $x=3$ and $y=1$ into the options.

For option C:

$$5(3) + 16(1) = 15 + 16 = 31$$

So option C passes through $(3,1)$.

For option D:

$$48(3) + 25(1) = 144 + 25 = 169$$

So option D also passes through $(3,1)$.

Hence passing through the midpoint alone is not sufficient. The correct chord equation must satisfy the standard midpoint formula for the ellipse, which gives

$$48x + 25y = 169$$

Therefore, the mathematically consistent answer is D, while the solution's marks C. This discrepancy should be noted.