Ask & Answer Question

Maths NCERT Exemplar Solutions Class 12th Chapter Six Linear Inequalities Ncert Solutions Maths class 11th Maths Class 11th

LE the common tangents to the curves $4 (x^{2} + y^{2}) = 9$ and y2 = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and $l$ respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $\frac{l}{e^{2}}$ is equal to

3 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Let y = mx + c is the common tangent

$s o c = \frac{1}{m} = \pm \frac{3}{2} \sqrt[]{1 + m^{2}} \Rightarrow m^{2} = \frac{1}{3}$

so equation of common tangents will be

$y = \pm \frac{1}{\sqrt[]{3}} x \pm \sqrt[]{3}$

which intersects at Q (3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^{2} = 1 - \frac{1}{4} = \frac{3}{4}$

and length of latus rectum

$= \frac{2 b^{2}}{a} = 3$

Hence

$\frac{l}{e^{2}} = \frac{3}{3 / 4} = 4$

0 Upvotes 0 Downvotes

Similar Questions for you

If one of the diameters of the circle $x^{2} + y^{2} - 2 \sqrt[]{2} x - 6 \sqrt[]{2} y + 14 = 0$ is a chord of circle. ${(x - 2 \sqrt[]{2})}^{2} + {(y - 2 \sqrt[]{2})}^{2} = r^{2},$ then the value of r² is equal to…………….

alok kumar singh

$x^{2} + y^{2} - 2 \sqrt[]{2} x - 6 \sqrt[]{2} y + 14 = 0$

centre $(\sqrt[]{2}, 3 \sqrt[]{2})$

radius ${({(\sqrt[]{2})}^{2} + {(3 \sqrt[]{2})}^{2} - 14)}^{1 / 2}$

$= {(2 + 18 - 14)}^{1 / 2} = (\sqrt[]{6})$

${(x - 2 \sqrt[]{2})}^{2} + {(y - 2 \sqrt[]{2})}^{2} = r^{2}$

centre $(2 \sqrt[]{2}, 2 \sqrt[]{2})$

OA = $\sqrt[]{{(2 \sqrt[]{2} - \sqrt[]{2})}^{2} + {(2 \sqrt[]{2} - 3 \sqrt[]{2})}^{2}} = \sqrt[]{2 + 2} = 2$

r² = ${(\sqrt[]{6})}^{2} + {(2)}^{2} = 6 + 4 = 10$

if $z^{2} + z + 1 = 0, z \in C, t h e n | \sum_{n = 1}^{15} {(z^{n} + {(- 1)}^{n} \frac{1}{z^{n}})}^{2} |$ $^{} {^{}^{} \frac{}{^{}}}^{}$ is equal to……….

Vishal Baghel

$z^{2} + z + 1 = 0 \Rightarrow z = w, w^{2}$

$| \sum_{n = 1}^{15} {(z^{n} + {(- 1)}^{n} \frac{1}{z^{n}})}^{2} | = | \sum_{n = 1}^{15} (z^{2 n} + \frac{1}{z^{2 n}} + 2 {(- 1)}^{n}) | = | \sum_{n = 1}^{15} w^{2 n} + \frac{1}{w^{2 n}} + 2 {(- 1)}^{n} |$

$= | \frac{w^{2} (1 - 1)}{1 - w^{2}} + \frac{\frac{1}{w^{2}} (1 - 1)}{1 - \frac{1}{w^{2}}} - 2 | = 2$

(i) If $- 4 x \geq 12$ , then $x \dots - 3$ .

(ii) If $\frac{3}{4} - x \leq - 3,$ then $x \dots 4$ .

(iii) If $\frac{2}{x + 2} > 0,$ then $x \dots - 2$ .

(iv) If $x > - 5$ , then $4 x \dots - 20$ .

(v) If $x > y$ and $z < 0$ , then $- x z \dots - y z$ .

(vi) If $p > 0$ and $q < 0$ , then $p - q \dots p$ .

(vii) If $x + 2 > 5$ , then $x \dots - 7$ or $x \dots 3$ .

(viii)If $- 2 x + 1 \geq 9$ , then $x \dots - 4$ .

Payal Gupta

This is a Fill in the blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) - 4 x \geq 12 \Rightarrow x \leq - 3 \\ H e n c e, t h e f i l l e r i s (\leq) \\ (i i) I f - \frac{3}{4} x \leq - 3 \Rightarrow x \geq 3 x \times \frac{4}{3} \Rightarrow x \geq 4 \\ H e n c e, t h e f i l l e r i s (\geq) \\ (i i i) I f \frac{2}{x + 2} > 0 \Rightarrow x > - 2 \\ H e n c e, t h e f i l l e r i s (>) \\ (i v) I f x > - 5 \Rightarrow 4 x > - 20 \\ H e n c e, t h e f i l l e r i s (>) \\ (v) I f x > y a n d z < 0, t h e n \\ x z < y z \Rightarrow - x z > - y z \\ H e n c e, t h e f i l l e r i s (>) \\ (v i) I f p > 0 a n d q < 0, t h e n \\ p - q > p \\ H e n c e, t h e f i l l e r i s (>) \\ (v i i) I f | x + 2 | > 5 t h e n \\ x + 2 < - 5 o r x + 2 > 5 \\ \Rightarrow x < - 5 - 2 o r x > 5 - 2 \\ \Rightarrow x < - 7 o r x > 3 \\ S o, x \in (- \infty, - 7) \cup (3, \infty) \\ H e n c e, t h e f i l l e r i s (<) o r (>) \\ (v i i i) I f - 2 x + 1 \geq 9 t h e n \\ \Rightarrow - 2 x \geq 9 - 1 \Rightarrow - 2 x \geq 8 \Rightarrow 2 x \leq - 8 \Rightarrow x \leq - 4 \\ H e n c e, t h e f i l l e r i s (\leq) \end{array}$

(i) If $x < y$ and $b < 0$ , then $\frac{x}{b} > \frac{y}{b} .$

(ii) If $x y > 0$ , then $x > 0$ and $y < 0$ .

(iii) If $x y > 0$ , then $x < 0$ and $y < 0$ .

(iv) If $x y < 0$ , then $x < 0$ and $y < 0$ .

(v) If $x < - 5$ and $x < - 2$ , then $x \in (- \infty, - 5)$ .

(vi) If $x < - 5$ and $x > 2$ , then $x \in (- 5, 2)$ .

(vii) If $x > - 2$ and $x < 9$ , then $x \in (- 2, 9)$ .

(viii) If $? x ? > 5$ , then $x \in (- \infty, - 5) \cup (5, \infty) .$

(ix) If $? x ? \leq 4$ , then $x \in [- 4, 4] .$

(x) Graph of $x < 3$ corresponds to Fig 6.12.

(xi) Graph of $x \geq 0$ corresponds to Fig 6.13.

(xii) Graph of $y \leq 0$ corresponds to Fig 6.14.

(xiii) Solution set of $x \geq 0$ and $y \leq 0$ corresponds to Fig 6.15.

(xiv) Solution set of $x \geq 0$ and $y \leq 1$ corresponds to Fig 6.16.

(xv) Solution set of $x + y \geq 0$ corresponds to Fig 6.17.

Payal Gupta

This is a True or False Type Questions as classified in NCERT Exemplar

$\begin{array}{l} (i) I f x < y a n d b < 0 \\ \Rightarrow \frac{x}{b} > \frac{y}{b} \\ H e n c e, s t a t e m e n t (i) i s F a l s e . \\ (i i) I f x, y > 0 t h e n x > 0, y > 0 o r x < 0, y < 0 \\ H e n c e, s t a t e m e n t (i i) i s F a l s e . \\ (i i i) I f x y > 0 t h e n x < 0, a n d y < 0 \\ H e n c e, s t a t e m e n t (i i i) i s T r u e . \\ (i v) I f x y < 0 t h e n x < 0, y > 0 o r x > 0, y < 0 \\ H e n c e, s t a t e m e n t (i v) i s F a l s e . \\ (v) I f x < - 5 a n d x < - 2 \Rightarrow x \in (- \infty, - 5) \\ H e n c e, s t a t e m e n t (v) i s T r u e . \\ (v i) I f x < - 5 a n d x > 2 t h e n x h a s n o v a l u e . \\ H e n c e, s t a t e m e n t (v i) i s F a l s e . \\ (v i i) I f x > - 2 a n d x < 9 \Rightarrow x \in (- 2, 9) \\ H e n c e, s t a t e m e n t (v i i) i s T r u e . \\ (v i i i) I f | x | > 5 t h e n x < - 5 o r x > 5 \\ \Rightarrow x \in (- \infty, - 5) \cup (5, \infty) \\ H e n c e, s t a t e m e n t (v i i i) i s F a l s e . \\ (i x) I f | x | \leq 4 t h e n - 4 \leq x \leq 4 \\ \Rightarrow x \in [- 4, 4] \\ H e n c e, s t a t e m e n t (i x) i s T r u e . \\ (x) T h e g i v e n g r a p h r e p r e s e n t s x \leq 3 \\ H e n c e, s t a t e m e n t (x) i s F a l s e . \\ (x i) T h e g i v e n g r a p h r e p r e s e n t s x \geq 0 \\ H e n c e, s t a t e m e n t (x i) i s T r u e . \\ (x i i) T h e g i v e n g r a p h r e p r e s e n t s y \geq 0 \\ H e n c e, s t a t e m e n t (x i i) i s &thins \end{array}$

30. Kindly consider the following y + 8 ≥ 2x

Payal Gupta

30.

30. For inequality y+8 ≤ 2x, the equation of the line is y+8=2x. We consider the table below to plot of y+8=2x.

$\begin{array}{l} x \\ y \end{array} | \begin{array}{l} 0 \\ - 8 \end{array} | \begin{array}{l} 4 \\ 0 \end{array} |$

The line devides the xy-plane into half planer I and II. We select a point (0,0) and check the correctness of the inequality.

i.e., 0+8 ≤ 2 × 0

0 ≤ 0 which is true.

So, the solution region is I which includes the rigin (0,0). The continuous line also indicates that any points on the line also satisfy the given inequality.