120275
If the curve \(y=a x^2+b x+c, x \in R\), passes through the point \((1,2)\) and the tangent line to this curve at origin is \(y=x\), then the possible values of \(a, b, c\) are
C Given, \(y^2=a^2+b x+c, x \in R\) Passes through the Point \((1,2)\) and the tangent line to this curve at origin \(y=x\).
As the curve passes through \((1,2)\) so the curve satisfies the point.
\(2=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Now tangent to the curve can be found by differentiating the curve.
\(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{ax}+\mathrm{b}\)
Now, comparing \(\mathrm{y}=\mathrm{x}\) with \(\mathrm{y}=\mathrm{mx}=\mathrm{c}, \mathrm{m}\) is slope we get,
\(\mathrm{m}=1\)
\(\left|\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(0,0)}\)
\(\mathrm{b}=1=1\)
Substituting the value of equation (ii) in equation (i),
\(a+c=1\)
Since, curve passes through the origin,
\(c=0\)
\(a=1\)Hence, \(\mathrm{a}=1, \mathrm{~b}=1, \mathrm{c}=0\)
JEE Main 24.02.2021
Parabola
120276
If the three normals drawn to the parabola, \(y^2\) \(=2 \mathrm{x}\) pass through the point \((\mathrm{a}, 0), \mathbf{a} \neq 0\), then \(\mathrm{a}\) must be greater than
1 \(\frac{1}{2}\)
2 \(\frac{1}{2}\)
3 -1
4 1
Explanation:
D Given,
\(y^2=2 x\)
Let, the equation of the normal
\(y=m x-2 a m-a m^3\)
From equation (i)
\(4 \mathrm{a}=2\)
\(\mathrm{a}=\frac{1}{2}\)
From equation (ii) and (iii), we get -
\(\mathrm{y}=\mathrm{mx}-\mathrm{m}-\frac{1}{2} \mathrm{~m}^3\)
It passes through the point \((\mathrm{a}, 0)\)
So,
\(\mathrm{m}(\mathrm{a})-\mathrm{m}-\frac{1}{2}(\mathrm{~m})^3=0\)
\(\mathrm{~m}\left(\mathrm{a}-1-\frac{\mathrm{m}^2}{2}\right)=0\)
\(\mathrm{~m}=0, \quad \mathrm{a}-1-\frac{\mathrm{m}^2}{2}=0\)
\(\mathrm{~m}=0, \quad \mathrm{~m}^2=2(\mathrm{a}-1)\)
As,
\(m^2>0\)
\(2(a-1)>0\)
\(a-1>0\)
\(a>1\)Hence, a must be greater than one.
JEE Main 16.03.2021
Parabola
120277
Let \(\mathbf{C}\) be the locus of the mirror image of a point on the parabola \(y^2=4 x\) with respect to the line \(y=x\). Then, the equation of tangent to \(\mathbf{C}\) at \(\mathbf{P}(\mathbf{2}, \mathbf{1})\) is
1 \(x-y=1\)
2 \(2 x+y=5\)
3 \(x+3 y=5\)
4 \(x+2 y=5\)
Explanation:
A Given,
\(y^2=4 x\)
Mirror image of the parabola \(y=x\)
\(C: x^2 =4 y\)
\(2 x =4 \cdot \frac{d y}{d x}\)
\(\frac{d y}{d x} =\frac{x}{2}\)
\(\left|\frac{d y}{d x}\right|_{p(2,1)} =\frac{2}{2}=1\)
We know equation of tangent, \(y-y_1=m\left(x-x_1\right)\)
\(y-1=1(x-2)\)
\(y_1=1, x_1=2\)
Equation of tangent of \((2,1)\),
\(\mathrm{xx}_1=2\left(\mathrm{y}+\mathrm{y}_1\right)\)
\(2 \mathrm{x}=2(\mathrm{y}+1)\)
\(\mathrm{x}-\mathrm{y}=1\)
JEE Main 20.03.2021
Parabola
120278
Let the tangent to the parabola S. \(y^2=2 x\) at the point \(P(2,2)\) meet the \(X\)-axis at \(Q\) and normal at it meet the parabola \(S\) at the point \(R\). Then, the area (in square units) of \(\triangle \mathrm{PQR}\) is equal
1 \(\frac{25}{2}\)
2 \(\frac{35}{2}\)
3 \(\frac{15}{2}\)
4 25
Explanation:
A Tangent to \(\mathrm{y}^2=2 \mathrm{x}\) at \(\mathrm{p}(2,2)\) is \(\mathrm{T}=0\)
\(2 \mathrm{y}=\mathrm{x}+2\)
\(\therefore \quad \mathrm{Q}=(-2,0)\)
Normal at \(\mathrm{P}\),
\(y-2=-\frac{(2)}{2 \times \frac{1}{2}}(x-2)\)
\(y-2=-2(x-2)\)
\(y=6-2 x\)
Now solving with \(\mathrm{y}^2\)
\(\mathrm{y}^2=2 \mathrm{x}\)
\(=\mathrm{R}\left(\frac{9}{2},-3\right)\)
\(\therefore \quad\) Area \((\triangle \mathrm{PQ} R)=\frac{1}{2}\left|\begin{array}{ccc}2 & 2 & 1 \\ -2 & 0 & 1 \\ \frac{9}{2} & -3 & 1\end{array}\right|\)
\(=\frac{1}{2}\left|2(0+3)-2\left(-2-\frac{9}{2}\right)+1(6-0)\right|\)
\(=\frac{1}{2}(6+13+6)\)
\(=\frac{25}{2} \mathrm{Sq} \text { units. }\)
120275
If the curve \(y=a x^2+b x+c, x \in R\), passes through the point \((1,2)\) and the tangent line to this curve at origin is \(y=x\), then the possible values of \(a, b, c\) are
C Given, \(y^2=a^2+b x+c, x \in R\) Passes through the Point \((1,2)\) and the tangent line to this curve at origin \(y=x\).
As the curve passes through \((1,2)\) so the curve satisfies the point.
\(2=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Now tangent to the curve can be found by differentiating the curve.
\(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{ax}+\mathrm{b}\)
Now, comparing \(\mathrm{y}=\mathrm{x}\) with \(\mathrm{y}=\mathrm{mx}=\mathrm{c}, \mathrm{m}\) is slope we get,
\(\mathrm{m}=1\)
\(\left|\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(0,0)}\)
\(\mathrm{b}=1=1\)
Substituting the value of equation (ii) in equation (i),
\(a+c=1\)
Since, curve passes through the origin,
\(c=0\)
\(a=1\)Hence, \(\mathrm{a}=1, \mathrm{~b}=1, \mathrm{c}=0\)
JEE Main 24.02.2021
Parabola
120276
If the three normals drawn to the parabola, \(y^2\) \(=2 \mathrm{x}\) pass through the point \((\mathrm{a}, 0), \mathbf{a} \neq 0\), then \(\mathrm{a}\) must be greater than
1 \(\frac{1}{2}\)
2 \(\frac{1}{2}\)
3 -1
4 1
Explanation:
D Given,
\(y^2=2 x\)
Let, the equation of the normal
\(y=m x-2 a m-a m^3\)
From equation (i)
\(4 \mathrm{a}=2\)
\(\mathrm{a}=\frac{1}{2}\)
From equation (ii) and (iii), we get -
\(\mathrm{y}=\mathrm{mx}-\mathrm{m}-\frac{1}{2} \mathrm{~m}^3\)
It passes through the point \((\mathrm{a}, 0)\)
So,
\(\mathrm{m}(\mathrm{a})-\mathrm{m}-\frac{1}{2}(\mathrm{~m})^3=0\)
\(\mathrm{~m}\left(\mathrm{a}-1-\frac{\mathrm{m}^2}{2}\right)=0\)
\(\mathrm{~m}=0, \quad \mathrm{a}-1-\frac{\mathrm{m}^2}{2}=0\)
\(\mathrm{~m}=0, \quad \mathrm{~m}^2=2(\mathrm{a}-1)\)
As,
\(m^2>0\)
\(2(a-1)>0\)
\(a-1>0\)
\(a>1\)Hence, a must be greater than one.
JEE Main 16.03.2021
Parabola
120277
Let \(\mathbf{C}\) be the locus of the mirror image of a point on the parabola \(y^2=4 x\) with respect to the line \(y=x\). Then, the equation of tangent to \(\mathbf{C}\) at \(\mathbf{P}(\mathbf{2}, \mathbf{1})\) is
1 \(x-y=1\)
2 \(2 x+y=5\)
3 \(x+3 y=5\)
4 \(x+2 y=5\)
Explanation:
A Given,
\(y^2=4 x\)
Mirror image of the parabola \(y=x\)
\(C: x^2 =4 y\)
\(2 x =4 \cdot \frac{d y}{d x}\)
\(\frac{d y}{d x} =\frac{x}{2}\)
\(\left|\frac{d y}{d x}\right|_{p(2,1)} =\frac{2}{2}=1\)
We know equation of tangent, \(y-y_1=m\left(x-x_1\right)\)
\(y-1=1(x-2)\)
\(y_1=1, x_1=2\)
Equation of tangent of \((2,1)\),
\(\mathrm{xx}_1=2\left(\mathrm{y}+\mathrm{y}_1\right)\)
\(2 \mathrm{x}=2(\mathrm{y}+1)\)
\(\mathrm{x}-\mathrm{y}=1\)
JEE Main 20.03.2021
Parabola
120278
Let the tangent to the parabola S. \(y^2=2 x\) at the point \(P(2,2)\) meet the \(X\)-axis at \(Q\) and normal at it meet the parabola \(S\) at the point \(R\). Then, the area (in square units) of \(\triangle \mathrm{PQR}\) is equal
1 \(\frac{25}{2}\)
2 \(\frac{35}{2}\)
3 \(\frac{15}{2}\)
4 25
Explanation:
A Tangent to \(\mathrm{y}^2=2 \mathrm{x}\) at \(\mathrm{p}(2,2)\) is \(\mathrm{T}=0\)
\(2 \mathrm{y}=\mathrm{x}+2\)
\(\therefore \quad \mathrm{Q}=(-2,0)\)
Normal at \(\mathrm{P}\),
\(y-2=-\frac{(2)}{2 \times \frac{1}{2}}(x-2)\)
\(y-2=-2(x-2)\)
\(y=6-2 x\)
Now solving with \(\mathrm{y}^2\)
\(\mathrm{y}^2=2 \mathrm{x}\)
\(=\mathrm{R}\left(\frac{9}{2},-3\right)\)
\(\therefore \quad\) Area \((\triangle \mathrm{PQ} R)=\frac{1}{2}\left|\begin{array}{ccc}2 & 2 & 1 \\ -2 & 0 & 1 \\ \frac{9}{2} & -3 & 1\end{array}\right|\)
\(=\frac{1}{2}\left|2(0+3)-2\left(-2-\frac{9}{2}\right)+1(6-0)\right|\)
\(=\frac{1}{2}(6+13+6)\)
\(=\frac{25}{2} \mathrm{Sq} \text { units. }\)
120275
If the curve \(y=a x^2+b x+c, x \in R\), passes through the point \((1,2)\) and the tangent line to this curve at origin is \(y=x\), then the possible values of \(a, b, c\) are
C Given, \(y^2=a^2+b x+c, x \in R\) Passes through the Point \((1,2)\) and the tangent line to this curve at origin \(y=x\).
As the curve passes through \((1,2)\) so the curve satisfies the point.
\(2=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Now tangent to the curve can be found by differentiating the curve.
\(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{ax}+\mathrm{b}\)
Now, comparing \(\mathrm{y}=\mathrm{x}\) with \(\mathrm{y}=\mathrm{mx}=\mathrm{c}, \mathrm{m}\) is slope we get,
\(\mathrm{m}=1\)
\(\left|\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(0,0)}\)
\(\mathrm{b}=1=1\)
Substituting the value of equation (ii) in equation (i),
\(a+c=1\)
Since, curve passes through the origin,
\(c=0\)
\(a=1\)Hence, \(\mathrm{a}=1, \mathrm{~b}=1, \mathrm{c}=0\)
JEE Main 24.02.2021
Parabola
120276
If the three normals drawn to the parabola, \(y^2\) \(=2 \mathrm{x}\) pass through the point \((\mathrm{a}, 0), \mathbf{a} \neq 0\), then \(\mathrm{a}\) must be greater than
1 \(\frac{1}{2}\)
2 \(\frac{1}{2}\)
3 -1
4 1
Explanation:
D Given,
\(y^2=2 x\)
Let, the equation of the normal
\(y=m x-2 a m-a m^3\)
From equation (i)
\(4 \mathrm{a}=2\)
\(\mathrm{a}=\frac{1}{2}\)
From equation (ii) and (iii), we get -
\(\mathrm{y}=\mathrm{mx}-\mathrm{m}-\frac{1}{2} \mathrm{~m}^3\)
It passes through the point \((\mathrm{a}, 0)\)
So,
\(\mathrm{m}(\mathrm{a})-\mathrm{m}-\frac{1}{2}(\mathrm{~m})^3=0\)
\(\mathrm{~m}\left(\mathrm{a}-1-\frac{\mathrm{m}^2}{2}\right)=0\)
\(\mathrm{~m}=0, \quad \mathrm{a}-1-\frac{\mathrm{m}^2}{2}=0\)
\(\mathrm{~m}=0, \quad \mathrm{~m}^2=2(\mathrm{a}-1)\)
As,
\(m^2>0\)
\(2(a-1)>0\)
\(a-1>0\)
\(a>1\)Hence, a must be greater than one.
JEE Main 16.03.2021
Parabola
120277
Let \(\mathbf{C}\) be the locus of the mirror image of a point on the parabola \(y^2=4 x\) with respect to the line \(y=x\). Then, the equation of tangent to \(\mathbf{C}\) at \(\mathbf{P}(\mathbf{2}, \mathbf{1})\) is
1 \(x-y=1\)
2 \(2 x+y=5\)
3 \(x+3 y=5\)
4 \(x+2 y=5\)
Explanation:
A Given,
\(y^2=4 x\)
Mirror image of the parabola \(y=x\)
\(C: x^2 =4 y\)
\(2 x =4 \cdot \frac{d y}{d x}\)
\(\frac{d y}{d x} =\frac{x}{2}\)
\(\left|\frac{d y}{d x}\right|_{p(2,1)} =\frac{2}{2}=1\)
We know equation of tangent, \(y-y_1=m\left(x-x_1\right)\)
\(y-1=1(x-2)\)
\(y_1=1, x_1=2\)
Equation of tangent of \((2,1)\),
\(\mathrm{xx}_1=2\left(\mathrm{y}+\mathrm{y}_1\right)\)
\(2 \mathrm{x}=2(\mathrm{y}+1)\)
\(\mathrm{x}-\mathrm{y}=1\)
JEE Main 20.03.2021
Parabola
120278
Let the tangent to the parabola S. \(y^2=2 x\) at the point \(P(2,2)\) meet the \(X\)-axis at \(Q\) and normal at it meet the parabola \(S\) at the point \(R\). Then, the area (in square units) of \(\triangle \mathrm{PQR}\) is equal
1 \(\frac{25}{2}\)
2 \(\frac{35}{2}\)
3 \(\frac{15}{2}\)
4 25
Explanation:
A Tangent to \(\mathrm{y}^2=2 \mathrm{x}\) at \(\mathrm{p}(2,2)\) is \(\mathrm{T}=0\)
\(2 \mathrm{y}=\mathrm{x}+2\)
\(\therefore \quad \mathrm{Q}=(-2,0)\)
Normal at \(\mathrm{P}\),
\(y-2=-\frac{(2)}{2 \times \frac{1}{2}}(x-2)\)
\(y-2=-2(x-2)\)
\(y=6-2 x\)
Now solving with \(\mathrm{y}^2\)
\(\mathrm{y}^2=2 \mathrm{x}\)
\(=\mathrm{R}\left(\frac{9}{2},-3\right)\)
\(\therefore \quad\) Area \((\triangle \mathrm{PQ} R)=\frac{1}{2}\left|\begin{array}{ccc}2 & 2 & 1 \\ -2 & 0 & 1 \\ \frac{9}{2} & -3 & 1\end{array}\right|\)
\(=\frac{1}{2}\left|2(0+3)-2\left(-2-\frac{9}{2}\right)+1(6-0)\right|\)
\(=\frac{1}{2}(6+13+6)\)
\(=\frac{25}{2} \mathrm{Sq} \text { units. }\)
120275
If the curve \(y=a x^2+b x+c, x \in R\), passes through the point \((1,2)\) and the tangent line to this curve at origin is \(y=x\), then the possible values of \(a, b, c\) are
C Given, \(y^2=a^2+b x+c, x \in R\) Passes through the Point \((1,2)\) and the tangent line to this curve at origin \(y=x\).
As the curve passes through \((1,2)\) so the curve satisfies the point.
\(2=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Now tangent to the curve can be found by differentiating the curve.
\(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{ax}+\mathrm{b}\)
Now, comparing \(\mathrm{y}=\mathrm{x}\) with \(\mathrm{y}=\mathrm{mx}=\mathrm{c}, \mathrm{m}\) is slope we get,
\(\mathrm{m}=1\)
\(\left|\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(0,0)}\)
\(\mathrm{b}=1=1\)
Substituting the value of equation (ii) in equation (i),
\(a+c=1\)
Since, curve passes through the origin,
\(c=0\)
\(a=1\)Hence, \(\mathrm{a}=1, \mathrm{~b}=1, \mathrm{c}=0\)
JEE Main 24.02.2021
Parabola
120276
If the three normals drawn to the parabola, \(y^2\) \(=2 \mathrm{x}\) pass through the point \((\mathrm{a}, 0), \mathbf{a} \neq 0\), then \(\mathrm{a}\) must be greater than
1 \(\frac{1}{2}\)
2 \(\frac{1}{2}\)
3 -1
4 1
Explanation:
D Given,
\(y^2=2 x\)
Let, the equation of the normal
\(y=m x-2 a m-a m^3\)
From equation (i)
\(4 \mathrm{a}=2\)
\(\mathrm{a}=\frac{1}{2}\)
From equation (ii) and (iii), we get -
\(\mathrm{y}=\mathrm{mx}-\mathrm{m}-\frac{1}{2} \mathrm{~m}^3\)
It passes through the point \((\mathrm{a}, 0)\)
So,
\(\mathrm{m}(\mathrm{a})-\mathrm{m}-\frac{1}{2}(\mathrm{~m})^3=0\)
\(\mathrm{~m}\left(\mathrm{a}-1-\frac{\mathrm{m}^2}{2}\right)=0\)
\(\mathrm{~m}=0, \quad \mathrm{a}-1-\frac{\mathrm{m}^2}{2}=0\)
\(\mathrm{~m}=0, \quad \mathrm{~m}^2=2(\mathrm{a}-1)\)
As,
\(m^2>0\)
\(2(a-1)>0\)
\(a-1>0\)
\(a>1\)Hence, a must be greater than one.
JEE Main 16.03.2021
Parabola
120277
Let \(\mathbf{C}\) be the locus of the mirror image of a point on the parabola \(y^2=4 x\) with respect to the line \(y=x\). Then, the equation of tangent to \(\mathbf{C}\) at \(\mathbf{P}(\mathbf{2}, \mathbf{1})\) is
1 \(x-y=1\)
2 \(2 x+y=5\)
3 \(x+3 y=5\)
4 \(x+2 y=5\)
Explanation:
A Given,
\(y^2=4 x\)
Mirror image of the parabola \(y=x\)
\(C: x^2 =4 y\)
\(2 x =4 \cdot \frac{d y}{d x}\)
\(\frac{d y}{d x} =\frac{x}{2}\)
\(\left|\frac{d y}{d x}\right|_{p(2,1)} =\frac{2}{2}=1\)
We know equation of tangent, \(y-y_1=m\left(x-x_1\right)\)
\(y-1=1(x-2)\)
\(y_1=1, x_1=2\)
Equation of tangent of \((2,1)\),
\(\mathrm{xx}_1=2\left(\mathrm{y}+\mathrm{y}_1\right)\)
\(2 \mathrm{x}=2(\mathrm{y}+1)\)
\(\mathrm{x}-\mathrm{y}=1\)
JEE Main 20.03.2021
Parabola
120278
Let the tangent to the parabola S. \(y^2=2 x\) at the point \(P(2,2)\) meet the \(X\)-axis at \(Q\) and normal at it meet the parabola \(S\) at the point \(R\). Then, the area (in square units) of \(\triangle \mathrm{PQR}\) is equal
1 \(\frac{25}{2}\)
2 \(\frac{35}{2}\)
3 \(\frac{15}{2}\)
4 25
Explanation:
A Tangent to \(\mathrm{y}^2=2 \mathrm{x}\) at \(\mathrm{p}(2,2)\) is \(\mathrm{T}=0\)
\(2 \mathrm{y}=\mathrm{x}+2\)
\(\therefore \quad \mathrm{Q}=(-2,0)\)
Normal at \(\mathrm{P}\),
\(y-2=-\frac{(2)}{2 \times \frac{1}{2}}(x-2)\)
\(y-2=-2(x-2)\)
\(y=6-2 x\)
Now solving with \(\mathrm{y}^2\)
\(\mathrm{y}^2=2 \mathrm{x}\)
\(=\mathrm{R}\left(\frac{9}{2},-3\right)\)
\(\therefore \quad\) Area \((\triangle \mathrm{PQ} R)=\frac{1}{2}\left|\begin{array}{ccc}2 & 2 & 1 \\ -2 & 0 & 1 \\ \frac{9}{2} & -3 & 1\end{array}\right|\)
\(=\frac{1}{2}\left|2(0+3)-2\left(-2-\frac{9}{2}\right)+1(6-0)\right|\)
\(=\frac{1}{2}(6+13+6)\)
\(=\frac{25}{2} \mathrm{Sq} \text { units. }\)
