QBManager

Ellipse

120627 The conic represented by
$x = 2 (\cos t + \sin t), y = 5 (\cos t - \sin t)$ is

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

C Given equations,
$x = 2 (\cos t + \sin t)$
$\frac{x}{2} = \cos t + \sin t$
$y = 5 (\cos t - \sin t)$
$\frac{y}{5} = (\cos t - \sin t)$
Eliminating from $t$ (i) and (ii), we get -
$\frac{x^{2}}{4} + \frac{y^{2}}{25} = 2$
$\frac{x^{2}}{8} + \frac{y^{2}}{50} = 1$ Which is an ellipse.

VITEEE-2019

Ellipse

120628 The line $x = a t^{2}$ meets the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ in the real points, if

1

| t | < 2

2 a parabola

y

is real, if

1 - t^{2} \geq 0

i.e.,

| t | \leq 1

.

3

| t | > 1

4 None of these

Explanation:

B
Given that, $x = a t^{2}$ and ellipse equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Now, putting $x = {at}^{2}$ in equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{{(a t^{2})}^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$t^{4} + \frac{y^{2}}{b^{2}} = 1$
$y^{2} = b^{2} (1 - t^{4})$
$= b^{2} (1 + t^{2}) (1 - t^{2})$
i.e.
(b.) a parabola
$y$ is real, if $1 - t^{2} \geq 0$ i.e., $| t | \leq 1$ .

UPSEE-2014

Ellipse

120629 If $t$ is a parameter, then $x = a (\sin t - \cos t), y =$ $b (\sin t + \cos t)$ represents :

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

C We have $x = a (\sin t - \cos t)$
$y = b (\sin t + \cos t)$
$\frac{x^{2}}{a^{2}} = (\sin t - \cos t)^{2}$
and $\frac{y^{2}}{b^{2}} = (\sin t + \cos t)^{2}$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = (\sin t - \cos t)^{2} + (\sin t + \cos t)^{2}$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2$

AMU-2013

Ellipse

120630 The total number of points on the curve $x^{2} -$ $4 y^{2} = 1$ at which the tangents to the curve are to the line $x = 2 y$ is

1 0

2 1

3 2

4 4

Explanation:

A Given, the curve
$x^{2} - 4 y^{2} = 1$
Differentiating w.r.t $x$ , we get
$2 x - 8 y \frac{d y}{d x} = 0$
$= \frac{d y}{d x} = \frac{2 x}{8 y} = \frac{x}{4 y}$
Let a point $p (h, k)$ lies on the curve.
$∴ h^{2} - 4 k^{2} = 1 \dots \dots$
$and {\frac{dy}{dx} |}_{h, k} = \frac{h}{4 k}$
Given the tangent is parallel to the line $x = 2 y$ or $y =$
$\frac{1}{2} x$
$∴ \frac{h}{4 k} = \frac{1}{2}$
$∴ h = 2 k$
This should satisfy equation (i)
$∴ h^{2} - 4 k^{2} = 1$
$4 k^{2} - 4 k^{2} = 0 \neq 1$ So, we get no. point on the curve.

Assam CEE-2018

Ellipse

120627 The conic represented by
$x = 2 (\cos t + \sin t), y = 5 (\cos t - \sin t)$ is

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

C Given equations,
$x = 2 (\cos t + \sin t)$
$\frac{x}{2} = \cos t + \sin t$
$y = 5 (\cos t - \sin t)$
$\frac{y}{5} = (\cos t - \sin t)$
Eliminating from $t$ (i) and (ii), we get -
$\frac{x^{2}}{4} + \frac{y^{2}}{25} = 2$
$\frac{x^{2}}{8} + \frac{y^{2}}{50} = 1$ Which is an ellipse.

VITEEE-2019

Ellipse

120628 The line $x = a t^{2}$ meets the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ in the real points, if

1

| t | < 2

2 a parabola

y

is real, if

1 - t^{2} \geq 0

i.e.,

| t | \leq 1

.

3

| t | > 1

4 None of these

Explanation:

B
Given that, $x = a t^{2}$ and ellipse equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Now, putting $x = {at}^{2}$ in equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{{(a t^{2})}^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$t^{4} + \frac{y^{2}}{b^{2}} = 1$
$y^{2} = b^{2} (1 - t^{4})$
$= b^{2} (1 + t^{2}) (1 - t^{2})$
i.e.
(b.) a parabola
$y$ is real, if $1 - t^{2} \geq 0$ i.e., $| t | \leq 1$ .

UPSEE-2014

Ellipse

120629 If $t$ is a parameter, then $x = a (\sin t - \cos t), y =$ $b (\sin t + \cos t)$ represents :

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

C We have $x = a (\sin t - \cos t)$
$y = b (\sin t + \cos t)$
$\frac{x^{2}}{a^{2}} = (\sin t - \cos t)^{2}$
and $\frac{y^{2}}{b^{2}} = (\sin t + \cos t)^{2}$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = (\sin t - \cos t)^{2} + (\sin t + \cos t)^{2}$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2$

AMU-2013

Ellipse

120630 The total number of points on the curve $x^{2} -$ $4 y^{2} = 1$ at which the tangents to the curve are to the line $x = 2 y$ is

1 0

2 1

3 2

4 4

Explanation:

A Given, the curve
$x^{2} - 4 y^{2} = 1$
Differentiating w.r.t $x$ , we get
$2 x - 8 y \frac{d y}{d x} = 0$
$= \frac{d y}{d x} = \frac{2 x}{8 y} = \frac{x}{4 y}$
Let a point $p (h, k)$ lies on the curve.
$∴ h^{2} - 4 k^{2} = 1 \dots \dots$
$and {\frac{dy}{dx} |}_{h, k} = \frac{h}{4 k}$
Given the tangent is parallel to the line $x = 2 y$ or $y =$
$\frac{1}{2} x$
$∴ \frac{h}{4 k} = \frac{1}{2}$
$∴ h = 2 k$
This should satisfy equation (i)
$∴ h^{2} - 4 k^{2} = 1$
$4 k^{2} - 4 k^{2} = 0 \neq 1$ So, we get no. point on the curve.

Assam CEE-2018

Ellipse

120627 The conic represented by
$x = 2 (\cos t + \sin t), y = 5 (\cos t - \sin t)$ is

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

C Given equations,
$x = 2 (\cos t + \sin t)$
$\frac{x}{2} = \cos t + \sin t$
$y = 5 (\cos t - \sin t)$
$\frac{y}{5} = (\cos t - \sin t)$
Eliminating from $t$ (i) and (ii), we get -
$\frac{x^{2}}{4} + \frac{y^{2}}{25} = 2$
$\frac{x^{2}}{8} + \frac{y^{2}}{50} = 1$ Which is an ellipse.

VITEEE-2019

Ellipse

120628 The line $x = a t^{2}$ meets the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ in the real points, if

1

| t | < 2

2 a parabola

y

is real, if

1 - t^{2} \geq 0

i.e.,

| t | \leq 1

.

3

| t | > 1

4 None of these

Explanation:

B
Given that, $x = a t^{2}$ and ellipse equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Now, putting $x = {at}^{2}$ in equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{{(a t^{2})}^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$t^{4} + \frac{y^{2}}{b^{2}} = 1$
$y^{2} = b^{2} (1 - t^{4})$
$= b^{2} (1 + t^{2}) (1 - t^{2})$
i.e.
(b.) a parabola
$y$ is real, if $1 - t^{2} \geq 0$ i.e., $| t | \leq 1$ .

UPSEE-2014

Ellipse

120629 If $t$ is a parameter, then $x = a (\sin t - \cos t), y =$ $b (\sin t + \cos t)$ represents :

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

C We have $x = a (\sin t - \cos t)$
$y = b (\sin t + \cos t)$
$\frac{x^{2}}{a^{2}} = (\sin t - \cos t)^{2}$
and $\frac{y^{2}}{b^{2}} = (\sin t + \cos t)^{2}$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = (\sin t - \cos t)^{2} + (\sin t + \cos t)^{2}$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2$

AMU-2013

Ellipse

120630 The total number of points on the curve $x^{2} -$ $4 y^{2} = 1$ at which the tangents to the curve are to the line $x = 2 y$ is

1 0

2 1

3 2

4 4

Explanation:

A Given, the curve
$x^{2} - 4 y^{2} = 1$
Differentiating w.r.t $x$ , we get
$2 x - 8 y \frac{d y}{d x} = 0$
$= \frac{d y}{d x} = \frac{2 x}{8 y} = \frac{x}{4 y}$
Let a point $p (h, k)$ lies on the curve.
$∴ h^{2} - 4 k^{2} = 1 \dots \dots$
$and {\frac{dy}{dx} |}_{h, k} = \frac{h}{4 k}$
Given the tangent is parallel to the line $x = 2 y$ or $y =$
$\frac{1}{2} x$
$∴ \frac{h}{4 k} = \frac{1}{2}$
$∴ h = 2 k$
This should satisfy equation (i)
$∴ h^{2} - 4 k^{2} = 1$
$4 k^{2} - 4 k^{2} = 0 \neq 1$ So, we get no. point on the curve.

Assam CEE-2018

Ellipse

120627 The conic represented by
$x = 2 (\cos t + \sin t), y = 5 (\cos t - \sin t)$ is

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

C Given equations,
$x = 2 (\cos t + \sin t)$
$\frac{x}{2} = \cos t + \sin t$
$y = 5 (\cos t - \sin t)$
$\frac{y}{5} = (\cos t - \sin t)$
Eliminating from $t$ (i) and (ii), we get -
$\frac{x^{2}}{4} + \frac{y^{2}}{25} = 2$
$\frac{x^{2}}{8} + \frac{y^{2}}{50} = 1$ Which is an ellipse.

VITEEE-2019

Ellipse

120628 The line $x = a t^{2}$ meets the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ in the real points, if

1

| t | < 2

2 a parabola

y

is real, if

1 - t^{2} \geq 0

i.e.,

| t | \leq 1

.

3

| t | > 1

4 None of these

Explanation:

B
Given that, $x = a t^{2}$ and ellipse equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Now, putting $x = {at}^{2}$ in equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$\frac{{(a t^{2})}^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
$t^{4} + \frac{y^{2}}{b^{2}} = 1$
$y^{2} = b^{2} (1 - t^{4})$
$= b^{2} (1 + t^{2}) (1 - t^{2})$
i.e.
(b.) a parabola
$y$ is real, if $1 - t^{2} \geq 0$ i.e., $| t | \leq 1$ .

UPSEE-2014

Ellipse

120629 If $t$ is a parameter, then $x = a (\sin t - \cos t), y =$ $b (\sin t + \cos t)$ represents :

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

C We have $x = a (\sin t - \cos t)$
$y = b (\sin t + \cos t)$
$\frac{x^{2}}{a^{2}} = (\sin t - \cos t)^{2}$
and $\frac{y^{2}}{b^{2}} = (\sin t + \cos t)^{2}$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = (\sin t - \cos t)^{2} + (\sin t + \cos t)^{2}$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2$

AMU-2013

Ellipse

120630 The total number of points on the curve $x^{2} -$ $4 y^{2} = 1$ at which the tangents to the curve are to the line $x = 2 y$ is

1 0

2 1

3 2

4 4

Explanation:

A Given, the curve
$x^{2} - 4 y^{2} = 1$
Differentiating w.r.t $x$ , we get
$2 x - 8 y \frac{d y}{d x} = 0$
$= \frac{d y}{d x} = \frac{2 x}{8 y} = \frac{x}{4 y}$
Let a point $p (h, k)$ lies on the curve.
$∴ h^{2} - 4 k^{2} = 1 \dots \dots$
$and {\frac{dy}{dx} |}_{h, k} = \frac{h}{4 k}$
Given the tangent is parallel to the line $x = 2 y$ or $y =$
$\frac{1}{2} x$
$∴ \frac{h}{4 k} = \frac{1}{2}$
$∴ h = 2 k$
This should satisfy equation (i)
$∴ h^{2} - 4 k^{2} = 1$
$4 k^{2} - 4 k^{2} = 0 \neq 1$ So, we get no. point on the curve.

Assam CEE-2018