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Integral Values of t for Which a Chord-of-Contact Fixed Point Lies Inside an Ellipse | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
From every point of the line (t+2)(x+y)=1(t+2)(x+y) = 1, t≠−2t \ne -2, tangents are drawn to the ellipse 4x2+16y2=14x^{2} + 16y^{2} = 1, and the chord of contact passes through a fixed point. Find the number of integral values of tt for which this fixed point lies inside the ellipse.
Solution
Answer: 2
Step 1: From (t+2)(x+y)=1(t+2)(x+y) = 1 we get x+y=1t+2x + y = \dfrac{1}{t+2}, so a general point of the line is
P(h, 1t+2−h)P\left(h,\ \frac{1}{t+2} - h\right)
 with hh taking every real value. Step 2: The chord of contact of PP with respect to 4x2+16y2=14x^{2}+16y^{2}=1 is T=0T=0:
4xh+16y(1t+2−h)=14xh + 16y\left(\frac{1}{t+2} - h\right) = 1
 Step 3: Grouping the terms in hh,
h(4x−16y)+16yt+2−1=0h\left(4x - 16y\right) + \frac{16y}{t+2} - 1 = 0
 Step 4: A point lying on every one of these chords must satisfy this for all hh, so both parts vanish:
4x−16y=0 ⟹ x=4y;16yt+2=1 ⟹ y=t+2164x - 16y = 0 \ \Longrightarrow\ x = 4y ; \qquad \frac{16y}{t+2} = 1 \ \Longrightarrow\ y = \frac{t+2}{16}
 So the fixed point is
(t+24, t+216)\left(\frac{t+2}{4},\ \frac{t+2}{16}\right)
 Step 5: Inside the ellipse means S1<0S_1 < 0:
4(t+24)2+16(t+216)2−1<04\left(\frac{t+2}{4}\right)^{2} + 16\left(\frac{t+2}{16}\right)^{2} - 1 < 0
(t+2)24+(t+2)216<1 ⟹ 5(t+2)216<1\frac{(t+2)^{2}}{4} + \frac{(t+2)^{2}}{16} < 1 \ \Longrightarrow\ \frac{5(t+2)^{2}}{16} < 1
 Step 6:
(t+2)2<165 ⟹ ∣t+2∣<45≈1.789(t+2)^{2} < \frac{16}{5} \ \Longrightarrow\ \left|t+2\right| < \frac{4}{\sqrt5} \approx 1.789
−45<t+2<45 ⟹ −3.789<t<−0.211-\frac{4}{\sqrt5} < t+2 < \frac{4}{\sqrt5} \ \Longrightarrow\ -3.789 < t < -0.211
 Step 7: The integers in that interval are −3,−2,−1-3, -2, -1, and t=−2t = -2 is excluded by the question, leaving
t=−3 and t=−1t = -3 \text{ and } t = -1
 so the count is 22. Answer: 2.002.00.
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