If the line $\frac{x - 1}{15} = \frac{y + 1}{16} = \frac{z - 1}{n}$ Intersect the curve $6 x^{2} + 5 y^{2} = 1$ , $z = 0$ ; then $10 n^{2} - 20 n + k = 0$ where the value of $\frac{k}{1315}$ is

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Solution

Ggiven line is $\frac{x - 1}{15} = \frac{y + 1}{16} = \frac{z - 1}{n} = p$ (say)
So any point on the given line is $(15 p + 1, 16 p - 1, n p + 1)$
Now this point line on the curve $6 x^{2} + 5 y^{2} = 1, z = 0$
$\Rightarrow 6 (15 p + 1)^{2} + 5 (16 p - 1)^{2} = 1$ and $n p + 1 = 0$
Using these,
$6 (- 15 / n + 1)^{2} + 5 (16 / n + 1)^{2} = 1$
$\Rightarrow 6 (n - 15)^{2} + 5 (n + 16)^{2} = n^{2}$
$\Rightarrow 10 n^{2} - 20 n + 2630 = 0$
Hence, $k = 2630$

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If the line x−115=y+116=z−1n Intersect the curve 6x2+5y2=1, z=0; then 10n2−20n+k=0 where the value of k1315 is

Ggiven line is x−115=y+116=z−1n=p (say)So any point on the given line is (15p+1,16p−1,np+1)Now this point line on the curve 6x2+5y2=1,z=0⇒6(15p+1)2+5(16p−1)2=1 and np+1=0Using these,6(−15/n+1)2+5(16/n+1)2=1⇒6(n−15)2+5(n+16)2=n2⇒10n2−20n+2630=0Hence, k=2630

If the line $\frac{x - 1}{15} = \frac{y + 1}{16} = \frac{z - 1}{n}$ Intersect the curve $6 x^{2} + 5 y^{2} = 1$ , $z = 0$ ; then $10 n^{2} - 20 n + k = 0$ where the value of $\frac{k}{1315}$ is