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 \(|t|\lt 2\)
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 \(\mathrm{x}=\mathrm{at}^2\) in equation \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\left(a t^2\right)^2}{a^2}+\frac{y^2}{b^2}=1\)
\(t^4+\frac{y^2}{b^2}=1\)
\(y^2=b^2\left(1-t^4\right)\)
\(=b^2\left(1+t^2\right)\left(1-t^2\right)\)
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 \(\mathrm{x}=\mathrm{a}(\sin \mathrm{t}-\cos \mathrm{t})\)
\(y=b(\sin t+\cos t)\)
\(\frac{x^2}{a^2}=(\sin t-\cos t)^2\)
and \(\quad \frac{\mathrm{y}^2}{\mathrm{~b}^2}=(\sin \mathrm{t}+\cos \mathrm{t})^2\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=(\sin t-\cos t)^2+(\sin t+\cos t)^2\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=2\)
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 \(\mathrm{x}\), 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 \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) lies on the curve.
\(\therefore \mathrm{h}^2-4 \mathrm{k}^2=1 \ldots \ldots\)
\(\text { and }\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{h}, \mathrm{k}}=\frac{\mathrm{h}}{4 \mathrm{k}}\)
Given the tangent is parallel to the line \(x=2 y\) or \(y=\)
\(\frac{1}{2} \mathrm{x}\)
\(\therefore \frac{\mathrm{h}}{4 \mathrm{k}}=\frac{1}{2}\)
\(\therefore \mathrm{h}=2 \mathrm{k}\)
This should satisfy equation (i)
\(\therefore \mathrm{h}^2-4 \mathrm{k}^2 =1\)
\(4 \mathrm{k}^2-4 \mathrm{k}^2 =0 \neq 1\)So, we get no. point on the curve.
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 \(|t|\lt 2\)
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 \(\mathrm{x}=\mathrm{at}^2\) in equation \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\left(a t^2\right)^2}{a^2}+\frac{y^2}{b^2}=1\)
\(t^4+\frac{y^2}{b^2}=1\)
\(y^2=b^2\left(1-t^4\right)\)
\(=b^2\left(1+t^2\right)\left(1-t^2\right)\)
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 \(\mathrm{x}=\mathrm{a}(\sin \mathrm{t}-\cos \mathrm{t})\)
\(y=b(\sin t+\cos t)\)
\(\frac{x^2}{a^2}=(\sin t-\cos t)^2\)
and \(\quad \frac{\mathrm{y}^2}{\mathrm{~b}^2}=(\sin \mathrm{t}+\cos \mathrm{t})^2\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=(\sin t-\cos t)^2+(\sin t+\cos t)^2\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=2\)
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 \(\mathrm{x}\), 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 \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) lies on the curve.
\(\therefore \mathrm{h}^2-4 \mathrm{k}^2=1 \ldots \ldots\)
\(\text { and }\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{h}, \mathrm{k}}=\frac{\mathrm{h}}{4 \mathrm{k}}\)
Given the tangent is parallel to the line \(x=2 y\) or \(y=\)
\(\frac{1}{2} \mathrm{x}\)
\(\therefore \frac{\mathrm{h}}{4 \mathrm{k}}=\frac{1}{2}\)
\(\therefore \mathrm{h}=2 \mathrm{k}\)
This should satisfy equation (i)
\(\therefore \mathrm{h}^2-4 \mathrm{k}^2 =1\)
\(4 \mathrm{k}^2-4 \mathrm{k}^2 =0 \neq 1\)So, we get no. point on the curve.
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 \(|t|\lt 2\)
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 \(\mathrm{x}=\mathrm{at}^2\) in equation \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\left(a t^2\right)^2}{a^2}+\frac{y^2}{b^2}=1\)
\(t^4+\frac{y^2}{b^2}=1\)
\(y^2=b^2\left(1-t^4\right)\)
\(=b^2\left(1+t^2\right)\left(1-t^2\right)\)
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 \(\mathrm{x}=\mathrm{a}(\sin \mathrm{t}-\cos \mathrm{t})\)
\(y=b(\sin t+\cos t)\)
\(\frac{x^2}{a^2}=(\sin t-\cos t)^2\)
and \(\quad \frac{\mathrm{y}^2}{\mathrm{~b}^2}=(\sin \mathrm{t}+\cos \mathrm{t})^2\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=(\sin t-\cos t)^2+(\sin t+\cos t)^2\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=2\)
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 \(\mathrm{x}\), 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 \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) lies on the curve.
\(\therefore \mathrm{h}^2-4 \mathrm{k}^2=1 \ldots \ldots\)
\(\text { and }\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{h}, \mathrm{k}}=\frac{\mathrm{h}}{4 \mathrm{k}}\)
Given the tangent is parallel to the line \(x=2 y\) or \(y=\)
\(\frac{1}{2} \mathrm{x}\)
\(\therefore \frac{\mathrm{h}}{4 \mathrm{k}}=\frac{1}{2}\)
\(\therefore \mathrm{h}=2 \mathrm{k}\)
This should satisfy equation (i)
\(\therefore \mathrm{h}^2-4 \mathrm{k}^2 =1\)
\(4 \mathrm{k}^2-4 \mathrm{k}^2 =0 \neq 1\)So, we get no. point on the curve.
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 \(|t|\lt 2\)
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 \(\mathrm{x}=\mathrm{at}^2\) in equation \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
\(\frac{\left(a t^2\right)^2}{a^2}+\frac{y^2}{b^2}=1\)
\(t^4+\frac{y^2}{b^2}=1\)
\(y^2=b^2\left(1-t^4\right)\)
\(=b^2\left(1+t^2\right)\left(1-t^2\right)\)
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 \(\mathrm{x}=\mathrm{a}(\sin \mathrm{t}-\cos \mathrm{t})\)
\(y=b(\sin t+\cos t)\)
\(\frac{x^2}{a^2}=(\sin t-\cos t)^2\)
and \(\quad \frac{\mathrm{y}^2}{\mathrm{~b}^2}=(\sin \mathrm{t}+\cos \mathrm{t})^2\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=(\sin t-\cos t)^2+(\sin t+\cos t)^2\)
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=2\)
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 \(\mathrm{x}\), 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 \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) lies on the curve.
\(\therefore \mathrm{h}^2-4 \mathrm{k}^2=1 \ldots \ldots\)
\(\text { and }\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{\mathrm{h}, \mathrm{k}}=\frac{\mathrm{h}}{4 \mathrm{k}}\)
Given the tangent is parallel to the line \(x=2 y\) or \(y=\)
\(\frac{1}{2} \mathrm{x}\)
\(\therefore \frac{\mathrm{h}}{4 \mathrm{k}}=\frac{1}{2}\)
\(\therefore \mathrm{h}=2 \mathrm{k}\)
This should satisfy equation (i)
\(\therefore \mathrm{h}^2-4 \mathrm{k}^2 =1\)
\(4 \mathrm{k}^2-4 \mathrm{k}^2 =0 \neq 1\)So, we get no. point on the curve.
