NCERT Class 11 Maths

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Since the line $\frac{x}{5}cos\theta +\frac{y}{12}sin\theta =1$ is the equation of tangent to the ellipse $\frac{x^2}{5^2}+\frac{y^2}{12^2}=1$ at (5 cos q, 12 sin q). Hence option (A) is correct option.

Let $?\frac{a}{r}$ , a, ar are in G.P. $\therefore \frac{a}{r},$ 2a, ar are in A.P.

$2a-\frac{a}{r}=ar-2a\Rightarrow 2-\frac{1}{r}=r-2\Rightarrow r^2-4r+1=0\Rightarrow r=2+\sqrt[]{3}$

$A/q,a{r}^2=3r^2\Rightarrow a=3$

$\begin{array}{l}\therefore d=ar-2a=3\sqrt[]{3}\\ \therefore r^2-d=7+\sqrt[]{3}\end{array}$

Length of latus rectum = 4 (distance between vertex and focus) = 4 (S – R)

Length of perpendicular from origin to xcosec a - y sec a = k cot2 a and x sin a + ycos a = k sin 2a are

$p=kcot\alpha sin\alpha cos\alpha =\frac{k}{2}.\frac{cos2\alpha }{sin2\alpha }sin2\alpha =\frac{k}{2}cos2\alpha \Rightarrow \frac{2p}{k}=cos2\alpha$

and $q=ksin2\alpha \Rightarrow \frac{q}{k}=sin2\alpha$

on solving these two we get 4p² + q² = k²

$e^{4x}+2e^{3x}-e^x-6=0$

$e^x=t\in \left(0,\infty \right)$

$t^4+2t^3-t-6=0$

Let f (t) = t⁴ + 2t³ – t – 6

f' (t) = 4t³ + 6t² – 1f

Þf' (0) = -1, f' (+ $\infty )$ = + $\infty$

For t > 0

Þ f' (t) = 0 has only one root.

One solution of f (t) = 0 is possible