120081
A particle is moving in the xy-plane along a curve \(C\) passing through the point \((3,3)\). The tangent to the curve \(C\) at the point \(P\) meets the \(x\)-axis at \(Q\). If the \(y\)-axis bisects the segment \(P Q\), then \(C\) is a parabola with
1 length of latus rectum 3
2 length of latus rectum 6
3 focus \(\left(\frac{4}{3}, 0\right)\)
4 focus \(\left(0, \frac{3}{4}\right)\)
Explanation:
A Given,
point \((3,3)\) Assume, that \(P\) is \((x, y)\)
So,
\(Y-y=y^{\prime}(X-x)\)
If, \(\mathrm{y}=0\)
\(x=x-\frac{y}{y^{\prime}}\)
\(\mathrm{Q}=\left(\mathrm{x}-\frac{\mathrm{y}}{\mathrm{y}^{\prime}}, 0\right)\)
Midpoint of \(\mathrm{PQ}\) lies on \(\mathrm{y}\)-axis
So,
\(x-\frac{y}{y^{\prime}}+x=0\)
\(2 x=\frac{y}{y^{\prime}}\)
\(\frac{y^{\prime}}{y}=\frac{1}{2 x}\)
\(\Rightarrow y^{\prime}=\frac{y}{2 x}\)
\(\Rightarrow 2 \frac{d y}{y}=\frac{d x}{x}\)
Integrate both sides
\(2 \log y=\log x+\log c\)
\(\log y^2=\log x+\log c\)
\(\Rightarrow y^2=x c\)
If passes through \((3,3)\)
So, \(=3\)
Length of latus rectum \(=3\)
Hence,
\(\quad \text { Focus }=\left(\frac{3}{4}, 0\right)\)
Integrate both sides -
If passes through \((3,3)\)
So, \(\mathrm{c}=3\)
Length of latus rectum \(=3\)
Hence,
JEE Main-24.06.2022
Parabola
120082
The equation of the directrix of parabola \(y^2-x\) \(+4 y+5=0\) is
120083
If the point \((a, 2 a)\) is an interior point of the region bounded by the parabola \(y^2=16 x\) and the double ordinate through focus then \(\qquad\)
1 a \(\lt 4\)
2 \(0\lt \) a \(\lt 4\)
3 \(0\lt \) a \(\lt 2\)
4 \(a>4\)
Explanation:
B Given,
\(y^2=16 x\)
\(y^2=4 \times 4 \times x\)
\(a=4\)
\(\mathrm{y}_1^2-4 \mathrm{ax}_1\lt 0 \quad \Rightarrow \text { interior }\)
\(O(0,0)\) and \(A\) lie on the same side of line \(x=4\) (a, 2a)
\(\mathrm{a}\lt 4\)
\(\mathrm{y}^2-16 \mathrm{x}=0\)
\(4 a^2-16 a\lt 0\)
\(4 \mathrm{a}(\mathrm{a}-4)\lt 0\)
(a) \((\mathrm{a}-4)\lt 0\)
\(\mathrm{a} \in(0,4)\)
So,
\(0\lt \mathrm{a}\lt 4\)
APEAPCET- 23.08.2021
Parabola
120084
The coordinates of the focus of the parabola described parametrically by \(x=\mathbf{5 t}^2+\mathbf{2}, \mathbf{y}=\) \(10 t+4\) (where \(t\) is a parameter) are \(\qquad\)
1 \((7,4)\)
2 \((3,4)\)
3 \((3,-4)\)
4 \((-7,4)\)
Explanation:
A Given,
\(\mathrm{x}=5 \mathrm{t}^2+2, \mathrm{y}=10 \mathrm{t}+4\)
\(\mathrm{t}^2=\frac{\mathrm{x}-2}{5}, \frac{\mathrm{y}-4}{10}=\mathrm{t}\)
\(\frac{\mathrm{x}-2}{5}=\left(\frac{\mathrm{y}-4}{10}\right)^2 \Rightarrow(\mathrm{y}-4)^2=20(\mathrm{x}-2)\)
\(\mathrm{Y}^2=20 \mathrm{X} \text { where }\)
\(\mathrm{Y}=\mathrm{y}-4\)
\(\mathrm{X}=\mathrm{x}-2\)
Coordinate of focus are \((5,0)\)
\(\text { i.e } x-2 =5\)
\(x =7\)
\(y-4 =0\)
\(y =4\)Hence, the required Co- ordinates are \((7,4)\).
AP EAPCET-25.08.2021
Parabola
120085
The length of the latus rectum of the parabola \((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\) is
1 \(\frac{1}{5}\)
2 \(\frac{2}{5}\)
3 \(\frac{3}{5}\)
4 \(\frac{4}{5}\)
Explanation:
B Given,
\((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\)
Focus is \((2,3)\)
Directrix : \(3 \mathrm{x}-4 \mathrm{y}+7\)
Now, length of latus rectum is -
\(l=2 \times\left|\frac{3 \times 2-4 \times 3+7}{\sqrt{3^2+4^2}}\right|\)
\(l=2 \times \frac{1}{5}=\frac{2}{5}\)
120081
A particle is moving in the xy-plane along a curve \(C\) passing through the point \((3,3)\). The tangent to the curve \(C\) at the point \(P\) meets the \(x\)-axis at \(Q\). If the \(y\)-axis bisects the segment \(P Q\), then \(C\) is a parabola with
1 length of latus rectum 3
2 length of latus rectum 6
3 focus \(\left(\frac{4}{3}, 0\right)\)
4 focus \(\left(0, \frac{3}{4}\right)\)
Explanation:
A Given,
point \((3,3)\) Assume, that \(P\) is \((x, y)\)
So,
\(Y-y=y^{\prime}(X-x)\)
If, \(\mathrm{y}=0\)
\(x=x-\frac{y}{y^{\prime}}\)
\(\mathrm{Q}=\left(\mathrm{x}-\frac{\mathrm{y}}{\mathrm{y}^{\prime}}, 0\right)\)
Midpoint of \(\mathrm{PQ}\) lies on \(\mathrm{y}\)-axis
So,
\(x-\frac{y}{y^{\prime}}+x=0\)
\(2 x=\frac{y}{y^{\prime}}\)
\(\frac{y^{\prime}}{y}=\frac{1}{2 x}\)
\(\Rightarrow y^{\prime}=\frac{y}{2 x}\)
\(\Rightarrow 2 \frac{d y}{y}=\frac{d x}{x}\)
Integrate both sides
\(2 \log y=\log x+\log c\)
\(\log y^2=\log x+\log c\)
\(\Rightarrow y^2=x c\)
If passes through \((3,3)\)
So, \(=3\)
Length of latus rectum \(=3\)
Hence,
\(\quad \text { Focus }=\left(\frac{3}{4}, 0\right)\)
Integrate both sides -
If passes through \((3,3)\)
So, \(\mathrm{c}=3\)
Length of latus rectum \(=3\)
Hence,
JEE Main-24.06.2022
Parabola
120082
The equation of the directrix of parabola \(y^2-x\) \(+4 y+5=0\) is
120083
If the point \((a, 2 a)\) is an interior point of the region bounded by the parabola \(y^2=16 x\) and the double ordinate through focus then \(\qquad\)
1 a \(\lt 4\)
2 \(0\lt \) a \(\lt 4\)
3 \(0\lt \) a \(\lt 2\)
4 \(a>4\)
Explanation:
B Given,
\(y^2=16 x\)
\(y^2=4 \times 4 \times x\)
\(a=4\)
\(\mathrm{y}_1^2-4 \mathrm{ax}_1\lt 0 \quad \Rightarrow \text { interior }\)
\(O(0,0)\) and \(A\) lie on the same side of line \(x=4\) (a, 2a)
\(\mathrm{a}\lt 4\)
\(\mathrm{y}^2-16 \mathrm{x}=0\)
\(4 a^2-16 a\lt 0\)
\(4 \mathrm{a}(\mathrm{a}-4)\lt 0\)
(a) \((\mathrm{a}-4)\lt 0\)
\(\mathrm{a} \in(0,4)\)
So,
\(0\lt \mathrm{a}\lt 4\)
APEAPCET- 23.08.2021
Parabola
120084
The coordinates of the focus of the parabola described parametrically by \(x=\mathbf{5 t}^2+\mathbf{2}, \mathbf{y}=\) \(10 t+4\) (where \(t\) is a parameter) are \(\qquad\)
1 \((7,4)\)
2 \((3,4)\)
3 \((3,-4)\)
4 \((-7,4)\)
Explanation:
A Given,
\(\mathrm{x}=5 \mathrm{t}^2+2, \mathrm{y}=10 \mathrm{t}+4\)
\(\mathrm{t}^2=\frac{\mathrm{x}-2}{5}, \frac{\mathrm{y}-4}{10}=\mathrm{t}\)
\(\frac{\mathrm{x}-2}{5}=\left(\frac{\mathrm{y}-4}{10}\right)^2 \Rightarrow(\mathrm{y}-4)^2=20(\mathrm{x}-2)\)
\(\mathrm{Y}^2=20 \mathrm{X} \text { where }\)
\(\mathrm{Y}=\mathrm{y}-4\)
\(\mathrm{X}=\mathrm{x}-2\)
Coordinate of focus are \((5,0)\)
\(\text { i.e } x-2 =5\)
\(x =7\)
\(y-4 =0\)
\(y =4\)Hence, the required Co- ordinates are \((7,4)\).
AP EAPCET-25.08.2021
Parabola
120085
The length of the latus rectum of the parabola \((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\) is
1 \(\frac{1}{5}\)
2 \(\frac{2}{5}\)
3 \(\frac{3}{5}\)
4 \(\frac{4}{5}\)
Explanation:
B Given,
\((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\)
Focus is \((2,3)\)
Directrix : \(3 \mathrm{x}-4 \mathrm{y}+7\)
Now, length of latus rectum is -
\(l=2 \times\left|\frac{3 \times 2-4 \times 3+7}{\sqrt{3^2+4^2}}\right|\)
\(l=2 \times \frac{1}{5}=\frac{2}{5}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Parabola
120081
A particle is moving in the xy-plane along a curve \(C\) passing through the point \((3,3)\). The tangent to the curve \(C\) at the point \(P\) meets the \(x\)-axis at \(Q\). If the \(y\)-axis bisects the segment \(P Q\), then \(C\) is a parabola with
1 length of latus rectum 3
2 length of latus rectum 6
3 focus \(\left(\frac{4}{3}, 0\right)\)
4 focus \(\left(0, \frac{3}{4}\right)\)
Explanation:
A Given,
point \((3,3)\) Assume, that \(P\) is \((x, y)\)
So,
\(Y-y=y^{\prime}(X-x)\)
If, \(\mathrm{y}=0\)
\(x=x-\frac{y}{y^{\prime}}\)
\(\mathrm{Q}=\left(\mathrm{x}-\frac{\mathrm{y}}{\mathrm{y}^{\prime}}, 0\right)\)
Midpoint of \(\mathrm{PQ}\) lies on \(\mathrm{y}\)-axis
So,
\(x-\frac{y}{y^{\prime}}+x=0\)
\(2 x=\frac{y}{y^{\prime}}\)
\(\frac{y^{\prime}}{y}=\frac{1}{2 x}\)
\(\Rightarrow y^{\prime}=\frac{y}{2 x}\)
\(\Rightarrow 2 \frac{d y}{y}=\frac{d x}{x}\)
Integrate both sides
\(2 \log y=\log x+\log c\)
\(\log y^2=\log x+\log c\)
\(\Rightarrow y^2=x c\)
If passes through \((3,3)\)
So, \(=3\)
Length of latus rectum \(=3\)
Hence,
\(\quad \text { Focus }=\left(\frac{3}{4}, 0\right)\)
Integrate both sides -
If passes through \((3,3)\)
So, \(\mathrm{c}=3\)
Length of latus rectum \(=3\)
Hence,
JEE Main-24.06.2022
Parabola
120082
The equation of the directrix of parabola \(y^2-x\) \(+4 y+5=0\) is
120083
If the point \((a, 2 a)\) is an interior point of the region bounded by the parabola \(y^2=16 x\) and the double ordinate through focus then \(\qquad\)
1 a \(\lt 4\)
2 \(0\lt \) a \(\lt 4\)
3 \(0\lt \) a \(\lt 2\)
4 \(a>4\)
Explanation:
B Given,
\(y^2=16 x\)
\(y^2=4 \times 4 \times x\)
\(a=4\)
\(\mathrm{y}_1^2-4 \mathrm{ax}_1\lt 0 \quad \Rightarrow \text { interior }\)
\(O(0,0)\) and \(A\) lie on the same side of line \(x=4\) (a, 2a)
\(\mathrm{a}\lt 4\)
\(\mathrm{y}^2-16 \mathrm{x}=0\)
\(4 a^2-16 a\lt 0\)
\(4 \mathrm{a}(\mathrm{a}-4)\lt 0\)
(a) \((\mathrm{a}-4)\lt 0\)
\(\mathrm{a} \in(0,4)\)
So,
\(0\lt \mathrm{a}\lt 4\)
APEAPCET- 23.08.2021
Parabola
120084
The coordinates of the focus of the parabola described parametrically by \(x=\mathbf{5 t}^2+\mathbf{2}, \mathbf{y}=\) \(10 t+4\) (where \(t\) is a parameter) are \(\qquad\)
1 \((7,4)\)
2 \((3,4)\)
3 \((3,-4)\)
4 \((-7,4)\)
Explanation:
A Given,
\(\mathrm{x}=5 \mathrm{t}^2+2, \mathrm{y}=10 \mathrm{t}+4\)
\(\mathrm{t}^2=\frac{\mathrm{x}-2}{5}, \frac{\mathrm{y}-4}{10}=\mathrm{t}\)
\(\frac{\mathrm{x}-2}{5}=\left(\frac{\mathrm{y}-4}{10}\right)^2 \Rightarrow(\mathrm{y}-4)^2=20(\mathrm{x}-2)\)
\(\mathrm{Y}^2=20 \mathrm{X} \text { where }\)
\(\mathrm{Y}=\mathrm{y}-4\)
\(\mathrm{X}=\mathrm{x}-2\)
Coordinate of focus are \((5,0)\)
\(\text { i.e } x-2 =5\)
\(x =7\)
\(y-4 =0\)
\(y =4\)Hence, the required Co- ordinates are \((7,4)\).
AP EAPCET-25.08.2021
Parabola
120085
The length of the latus rectum of the parabola \((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\) is
1 \(\frac{1}{5}\)
2 \(\frac{2}{5}\)
3 \(\frac{3}{5}\)
4 \(\frac{4}{5}\)
Explanation:
B Given,
\((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\)
Focus is \((2,3)\)
Directrix : \(3 \mathrm{x}-4 \mathrm{y}+7\)
Now, length of latus rectum is -
\(l=2 \times\left|\frac{3 \times 2-4 \times 3+7}{\sqrt{3^2+4^2}}\right|\)
\(l=2 \times \frac{1}{5}=\frac{2}{5}\)
120081
A particle is moving in the xy-plane along a curve \(C\) passing through the point \((3,3)\). The tangent to the curve \(C\) at the point \(P\) meets the \(x\)-axis at \(Q\). If the \(y\)-axis bisects the segment \(P Q\), then \(C\) is a parabola with
1 length of latus rectum 3
2 length of latus rectum 6
3 focus \(\left(\frac{4}{3}, 0\right)\)
4 focus \(\left(0, \frac{3}{4}\right)\)
Explanation:
A Given,
point \((3,3)\) Assume, that \(P\) is \((x, y)\)
So,
\(Y-y=y^{\prime}(X-x)\)
If, \(\mathrm{y}=0\)
\(x=x-\frac{y}{y^{\prime}}\)
\(\mathrm{Q}=\left(\mathrm{x}-\frac{\mathrm{y}}{\mathrm{y}^{\prime}}, 0\right)\)
Midpoint of \(\mathrm{PQ}\) lies on \(\mathrm{y}\)-axis
So,
\(x-\frac{y}{y^{\prime}}+x=0\)
\(2 x=\frac{y}{y^{\prime}}\)
\(\frac{y^{\prime}}{y}=\frac{1}{2 x}\)
\(\Rightarrow y^{\prime}=\frac{y}{2 x}\)
\(\Rightarrow 2 \frac{d y}{y}=\frac{d x}{x}\)
Integrate both sides
\(2 \log y=\log x+\log c\)
\(\log y^2=\log x+\log c\)
\(\Rightarrow y^2=x c\)
If passes through \((3,3)\)
So, \(=3\)
Length of latus rectum \(=3\)
Hence,
\(\quad \text { Focus }=\left(\frac{3}{4}, 0\right)\)
Integrate both sides -
If passes through \((3,3)\)
So, \(\mathrm{c}=3\)
Length of latus rectum \(=3\)
Hence,
JEE Main-24.06.2022
Parabola
120082
The equation of the directrix of parabola \(y^2-x\) \(+4 y+5=0\) is
120083
If the point \((a, 2 a)\) is an interior point of the region bounded by the parabola \(y^2=16 x\) and the double ordinate through focus then \(\qquad\)
1 a \(\lt 4\)
2 \(0\lt \) a \(\lt 4\)
3 \(0\lt \) a \(\lt 2\)
4 \(a>4\)
Explanation:
B Given,
\(y^2=16 x\)
\(y^2=4 \times 4 \times x\)
\(a=4\)
\(\mathrm{y}_1^2-4 \mathrm{ax}_1\lt 0 \quad \Rightarrow \text { interior }\)
\(O(0,0)\) and \(A\) lie on the same side of line \(x=4\) (a, 2a)
\(\mathrm{a}\lt 4\)
\(\mathrm{y}^2-16 \mathrm{x}=0\)
\(4 a^2-16 a\lt 0\)
\(4 \mathrm{a}(\mathrm{a}-4)\lt 0\)
(a) \((\mathrm{a}-4)\lt 0\)
\(\mathrm{a} \in(0,4)\)
So,
\(0\lt \mathrm{a}\lt 4\)
APEAPCET- 23.08.2021
Parabola
120084
The coordinates of the focus of the parabola described parametrically by \(x=\mathbf{5 t}^2+\mathbf{2}, \mathbf{y}=\) \(10 t+4\) (where \(t\) is a parameter) are \(\qquad\)
1 \((7,4)\)
2 \((3,4)\)
3 \((3,-4)\)
4 \((-7,4)\)
Explanation:
A Given,
\(\mathrm{x}=5 \mathrm{t}^2+2, \mathrm{y}=10 \mathrm{t}+4\)
\(\mathrm{t}^2=\frac{\mathrm{x}-2}{5}, \frac{\mathrm{y}-4}{10}=\mathrm{t}\)
\(\frac{\mathrm{x}-2}{5}=\left(\frac{\mathrm{y}-4}{10}\right)^2 \Rightarrow(\mathrm{y}-4)^2=20(\mathrm{x}-2)\)
\(\mathrm{Y}^2=20 \mathrm{X} \text { where }\)
\(\mathrm{Y}=\mathrm{y}-4\)
\(\mathrm{X}=\mathrm{x}-2\)
Coordinate of focus are \((5,0)\)
\(\text { i.e } x-2 =5\)
\(x =7\)
\(y-4 =0\)
\(y =4\)Hence, the required Co- ordinates are \((7,4)\).
AP EAPCET-25.08.2021
Parabola
120085
The length of the latus rectum of the parabola \((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\) is
1 \(\frac{1}{5}\)
2 \(\frac{2}{5}\)
3 \(\frac{3}{5}\)
4 \(\frac{4}{5}\)
Explanation:
B Given,
\((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\)
Focus is \((2,3)\)
Directrix : \(3 \mathrm{x}-4 \mathrm{y}+7\)
Now, length of latus rectum is -
\(l=2 \times\left|\frac{3 \times 2-4 \times 3+7}{\sqrt{3^2+4^2}}\right|\)
\(l=2 \times \frac{1}{5}=\frac{2}{5}\)
120081
A particle is moving in the xy-plane along a curve \(C\) passing through the point \((3,3)\). The tangent to the curve \(C\) at the point \(P\) meets the \(x\)-axis at \(Q\). If the \(y\)-axis bisects the segment \(P Q\), then \(C\) is a parabola with
1 length of latus rectum 3
2 length of latus rectum 6
3 focus \(\left(\frac{4}{3}, 0\right)\)
4 focus \(\left(0, \frac{3}{4}\right)\)
Explanation:
A Given,
point \((3,3)\) Assume, that \(P\) is \((x, y)\)
So,
\(Y-y=y^{\prime}(X-x)\)
If, \(\mathrm{y}=0\)
\(x=x-\frac{y}{y^{\prime}}\)
\(\mathrm{Q}=\left(\mathrm{x}-\frac{\mathrm{y}}{\mathrm{y}^{\prime}}, 0\right)\)
Midpoint of \(\mathrm{PQ}\) lies on \(\mathrm{y}\)-axis
So,
\(x-\frac{y}{y^{\prime}}+x=0\)
\(2 x=\frac{y}{y^{\prime}}\)
\(\frac{y^{\prime}}{y}=\frac{1}{2 x}\)
\(\Rightarrow y^{\prime}=\frac{y}{2 x}\)
\(\Rightarrow 2 \frac{d y}{y}=\frac{d x}{x}\)
Integrate both sides
\(2 \log y=\log x+\log c\)
\(\log y^2=\log x+\log c\)
\(\Rightarrow y^2=x c\)
If passes through \((3,3)\)
So, \(=3\)
Length of latus rectum \(=3\)
Hence,
\(\quad \text { Focus }=\left(\frac{3}{4}, 0\right)\)
Integrate both sides -
If passes through \((3,3)\)
So, \(\mathrm{c}=3\)
Length of latus rectum \(=3\)
Hence,
JEE Main-24.06.2022
Parabola
120082
The equation of the directrix of parabola \(y^2-x\) \(+4 y+5=0\) is
120083
If the point \((a, 2 a)\) is an interior point of the region bounded by the parabola \(y^2=16 x\) and the double ordinate through focus then \(\qquad\)
1 a \(\lt 4\)
2 \(0\lt \) a \(\lt 4\)
3 \(0\lt \) a \(\lt 2\)
4 \(a>4\)
Explanation:
B Given,
\(y^2=16 x\)
\(y^2=4 \times 4 \times x\)
\(a=4\)
\(\mathrm{y}_1^2-4 \mathrm{ax}_1\lt 0 \quad \Rightarrow \text { interior }\)
\(O(0,0)\) and \(A\) lie on the same side of line \(x=4\) (a, 2a)
\(\mathrm{a}\lt 4\)
\(\mathrm{y}^2-16 \mathrm{x}=0\)
\(4 a^2-16 a\lt 0\)
\(4 \mathrm{a}(\mathrm{a}-4)\lt 0\)
(a) \((\mathrm{a}-4)\lt 0\)
\(\mathrm{a} \in(0,4)\)
So,
\(0\lt \mathrm{a}\lt 4\)
APEAPCET- 23.08.2021
Parabola
120084
The coordinates of the focus of the parabola described parametrically by \(x=\mathbf{5 t}^2+\mathbf{2}, \mathbf{y}=\) \(10 t+4\) (where \(t\) is a parameter) are \(\qquad\)
1 \((7,4)\)
2 \((3,4)\)
3 \((3,-4)\)
4 \((-7,4)\)
Explanation:
A Given,
\(\mathrm{x}=5 \mathrm{t}^2+2, \mathrm{y}=10 \mathrm{t}+4\)
\(\mathrm{t}^2=\frac{\mathrm{x}-2}{5}, \frac{\mathrm{y}-4}{10}=\mathrm{t}\)
\(\frac{\mathrm{x}-2}{5}=\left(\frac{\mathrm{y}-4}{10}\right)^2 \Rightarrow(\mathrm{y}-4)^2=20(\mathrm{x}-2)\)
\(\mathrm{Y}^2=20 \mathrm{X} \text { where }\)
\(\mathrm{Y}=\mathrm{y}-4\)
\(\mathrm{X}=\mathrm{x}-2\)
Coordinate of focus are \((5,0)\)
\(\text { i.e } x-2 =5\)
\(x =7\)
\(y-4 =0\)
\(y =4\)Hence, the required Co- ordinates are \((7,4)\).
AP EAPCET-25.08.2021
Parabola
120085
The length of the latus rectum of the parabola \((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\) is
1 \(\frac{1}{5}\)
2 \(\frac{2}{5}\)
3 \(\frac{3}{5}\)
4 \(\frac{4}{5}\)
Explanation:
B Given,
\((x-2)^2+(y-3)^2=\frac{1}{25}(3 x-4 y+7)^2\)
Focus is \((2,3)\)
Directrix : \(3 \mathrm{x}-4 \mathrm{y}+7\)
Now, length of latus rectum is -
\(l=2 \times\left|\frac{3 \times 2-4 \times 3+7}{\sqrt{3^2+4^2}}\right|\)
\(l=2 \times \frac{1}{5}=\frac{2}{5}\)
