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Maths Question 11 – JEE-MAIN 2025

Let the point P of the focal chord PQ of the parabola y2=16x be (1,−4). If the focus of the parabola divides the chord PQ in the ratio m:n, gcd(m,n)=1, then m2+n2 is equal to :

Identify the standard form of the parabola y2=4ax and determine the coordinates of its focus (a,0).

Step 1: Identify Parabola Parameters and Focus✦ Active

The given parabola equation is y2=16x. Comparing this with the standard form y2=4ax, we find 4a=16, which implies a=4. Therefore, the focus of the parabola, S, is at (a,0)=(4,0).

Step 2: Find Parametric Coordinates of P and Q○ Expand

Let the parametric coordinates of point P be (at12,2at1). Given P=(1,−4) and a=4, we have 4t12=1⟹t12=14⟹t1=±12. Also, 2(4)t1=−4⟹8t1=−4⟹t1=−12. Thus, t1=−12. For a focal chord, the product of the parameters is t1t2=−1. Substituting t1=−12, we get (−12)t2=−1⟹t2=2. The coordinates of point Q are (at22,2at2)=(4(22),2(4)(2))=(16,16).

💡 Teacher's Secret Hint

Remember the condition t1t2=−1 for a focal chord passing through the focus.

Step 3: Apply Section Formula and Calculate○ Expand

The focus S(4,0) divides the chord PQ, with P(1,−4) and Q(16,16), in the ratio m:n. Using the section formula for the x-coordinate:

4=m(16)+n(1)m+n

This simplifies to 4(m+n)=16m+n⟹4m+4n=16m+n⟹3n=12m⟹n=4m. The ratio m:n=1:4. Since gcd(m,n)=1, we have m=1 and n=4. Finally, we need to calculate m2+n2:

m2+n2=12+42=1+16=17
💡 Teacher's Secret Hint

Ensure the ratio m:n is in its simplest form as gcd(m,n)=1 is given.

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