NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120810
The equation of a common tangent to the curves, \(y^2=16 x\) and \(x y=-4\) is
1 \(x-y+4=0\)
2 \(x+y+4=0\)
3 \(x-2 y+16=0\)
4 \(2 x-y+2=0\)
Explanation:
A Given,
\(y=m x+\frac{4}{m}\) is always tangent to
\(y^2=16 x\)
If it is tangent to the \(x y=-4\)
\(x\left(m x+\frac{4}{m}\right)=-4\)
\(m^2 x^2+4 x=-4 m\)
\(m^2 x^2+4 x+4 m=0\)
For tangent \(\mathrm{D}=0\)
\(16-16 \mathrm{~m}^3=0 \Rightarrow \mathrm{m}=1 \text { put in equation (i) }\)
\(y=x+4\)
\(x-y+4=0\)
JEE Main 12.04.2019
Hyperbola
120811
Let \(P(3,3)\) be a point on the hyperbola, \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\). If the normal to it at \(P\) intersects the \(X\)-axis at \((9,0)\) and \(e\) is its eccentricity, then the ordered pair \(\left(a^2, \mathrm{e}^2\right)\) is equal to
1 \(\left(\frac{9}{2}, 3\right)\)
2 \(\left(\frac{3}{2}, 2\right)\)
3 \(\left(\frac{9}{2}, 2\right)\)
4 \((9,3)\)
Explanation:
A Given,
According to question,
\(\mathrm{p}(3,3) \text { lies on } \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
So, \(\quad \frac{9}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=1\)
Normal at \((3,3)\) is \(\rightarrow\)
\(y-3=-\frac{a^2}{b^2}(x-3)\)
Passes through \((9,0) \rightarrow\)
\(\mathrm{b}^2=2 \mathrm{a}^2\)
So,
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=3\)
Also,
\(\mathrm{a}^2=\frac{9}{2} \quad\) \{from equation (i) \& equation (ii) \(\}\)
Thus, \(\quad\left(\mathrm{a}^2, \mathrm{e}^2\right)=\left(\frac{9}{2}, 3\right)\)
JEE Main 04.09.2020
Hyperbola
120812
If the line \(y=m x+c\) is common tangent to the hyperbola \(\frac{x^2}{100}-\frac{y^2}{64}=1\) and the circle \(x^2+y^2=\) 36 , then which one of the following is true?
1 \(\mathrm{c}^2=369\)
2 \(5 \mathrm{~m}=4\)
3 \(4 \mathrm{c}^2=369\)
4 \(8 \mathrm{~m}+5=0\)
Explanation:
C Given,
Line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) is common tangent to the hyperbola
\(\frac{x^2}{100}-\frac{y^2}{64}=1\) and the circle \(x^2+y^2=36\)
We know that, for hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(c=\sqrt{a^2 m^2-b^2}\)
\(c=\sqrt{100 m^2-64}\)
For the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2, \mathrm{c}\) is given by \(\rightarrow\)
\(\mathrm{c}=\mathrm{a} \sqrt{1+\mathrm{m}^2}\)
\(\mathrm{c}=6 \sqrt{1+\mathrm{m}^2}\)
So, from equation (i) and equation (ii), we get -
\(\sqrt{100 \mathrm{~m}^2-64}=6 \sqrt{1+\mathrm{m}^2}\)
Square both side,
\(100 \mathrm{~m}^2-64=36\left(1+\mathrm{m}^2\right)\)
\(64 \mathrm{~m}^2=100\)
\(\mathrm{~m}= \pm \frac{10}{8}\)
From equation (ii) \(\rightarrow\)
\(c=6 \sqrt{1+\frac{100}{64}}\)
\(c=6 \sqrt{\frac{64+100}{64}}\)
\(c=6\left(\frac{1}{8} \sqrt{164}\right)\)
Taking square both side -
\(\mathrm{c}^2=36 \times\left(\frac{164}{64}\right)\)
\(\mathrm{c}^2=\frac{369}{4}\)
\(4 \mathrm{c}^2=369\)
JEE Main 05.09.2020
Hyperbola
120813
If a hyperbola passes through the point \(P(10\), \(16)\) and it has vertices at \(( \pm 6,0)\) then the equation of the normal to it at \(P\) is
1 \(3 x+4 y=94\)
2 \(x+2 y=42\)
3 \(2 x+5 y=100\)
4 \(x+3 y=58\)
Explanation:
C Given,
Let equation of hyperbola is -
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Given, its vertex at \(( \pm 6,0)\), so, \(\mathrm{a}=6\)
Also passes through \((10,16)\) so, \(\frac{100}{36}-\frac{256}{\mathrm{~b}^2}=1\)
\(b^2=144\)
\(\therefore\) Equation \(\frac{\mathrm{x}^2}{36}-\frac{\mathrm{y}^2}{144}=1\)
Equation of normal is -
\(\frac{a^2 x}{x_1}+\frac{b^2}{y_1} y=a^2+b^2\)
\(\mathrm{x}_1=10, \mathrm{y}_1=16\)
\(\frac{36 x}{10}+\frac{144 y}{16}=144+36\)
\(\therefore \quad 2 x+5 y=100\)
(from solving)
120810
The equation of a common tangent to the curves, \(y^2=16 x\) and \(x y=-4\) is
1 \(x-y+4=0\)
2 \(x+y+4=0\)
3 \(x-2 y+16=0\)
4 \(2 x-y+2=0\)
Explanation:
A Given,
\(y=m x+\frac{4}{m}\) is always tangent to
\(y^2=16 x\)
If it is tangent to the \(x y=-4\)
\(x\left(m x+\frac{4}{m}\right)=-4\)
\(m^2 x^2+4 x=-4 m\)
\(m^2 x^2+4 x+4 m=0\)
For tangent \(\mathrm{D}=0\)
\(16-16 \mathrm{~m}^3=0 \Rightarrow \mathrm{m}=1 \text { put in equation (i) }\)
\(y=x+4\)
\(x-y+4=0\)
JEE Main 12.04.2019
Hyperbola
120811
Let \(P(3,3)\) be a point on the hyperbola, \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\). If the normal to it at \(P\) intersects the \(X\)-axis at \((9,0)\) and \(e\) is its eccentricity, then the ordered pair \(\left(a^2, \mathrm{e}^2\right)\) is equal to
1 \(\left(\frac{9}{2}, 3\right)\)
2 \(\left(\frac{3}{2}, 2\right)\)
3 \(\left(\frac{9}{2}, 2\right)\)
4 \((9,3)\)
Explanation:
A Given,
According to question,
\(\mathrm{p}(3,3) \text { lies on } \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
So, \(\quad \frac{9}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=1\)
Normal at \((3,3)\) is \(\rightarrow\)
\(y-3=-\frac{a^2}{b^2}(x-3)\)
Passes through \((9,0) \rightarrow\)
\(\mathrm{b}^2=2 \mathrm{a}^2\)
So,
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=3\)
Also,
\(\mathrm{a}^2=\frac{9}{2} \quad\) \{from equation (i) \& equation (ii) \(\}\)
Thus, \(\quad\left(\mathrm{a}^2, \mathrm{e}^2\right)=\left(\frac{9}{2}, 3\right)\)
JEE Main 04.09.2020
Hyperbola
120812
If the line \(y=m x+c\) is common tangent to the hyperbola \(\frac{x^2}{100}-\frac{y^2}{64}=1\) and the circle \(x^2+y^2=\) 36 , then which one of the following is true?
1 \(\mathrm{c}^2=369\)
2 \(5 \mathrm{~m}=4\)
3 \(4 \mathrm{c}^2=369\)
4 \(8 \mathrm{~m}+5=0\)
Explanation:
C Given,
Line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) is common tangent to the hyperbola
\(\frac{x^2}{100}-\frac{y^2}{64}=1\) and the circle \(x^2+y^2=36\)
We know that, for hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(c=\sqrt{a^2 m^2-b^2}\)
\(c=\sqrt{100 m^2-64}\)
For the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2, \mathrm{c}\) is given by \(\rightarrow\)
\(\mathrm{c}=\mathrm{a} \sqrt{1+\mathrm{m}^2}\)
\(\mathrm{c}=6 \sqrt{1+\mathrm{m}^2}\)
So, from equation (i) and equation (ii), we get -
\(\sqrt{100 \mathrm{~m}^2-64}=6 \sqrt{1+\mathrm{m}^2}\)
Square both side,
\(100 \mathrm{~m}^2-64=36\left(1+\mathrm{m}^2\right)\)
\(64 \mathrm{~m}^2=100\)
\(\mathrm{~m}= \pm \frac{10}{8}\)
From equation (ii) \(\rightarrow\)
\(c=6 \sqrt{1+\frac{100}{64}}\)
\(c=6 \sqrt{\frac{64+100}{64}}\)
\(c=6\left(\frac{1}{8} \sqrt{164}\right)\)
Taking square both side -
\(\mathrm{c}^2=36 \times\left(\frac{164}{64}\right)\)
\(\mathrm{c}^2=\frac{369}{4}\)
\(4 \mathrm{c}^2=369\)
JEE Main 05.09.2020
Hyperbola
120813
If a hyperbola passes through the point \(P(10\), \(16)\) and it has vertices at \(( \pm 6,0)\) then the equation of the normal to it at \(P\) is
1 \(3 x+4 y=94\)
2 \(x+2 y=42\)
3 \(2 x+5 y=100\)
4 \(x+3 y=58\)
Explanation:
C Given,
Let equation of hyperbola is -
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Given, its vertex at \(( \pm 6,0)\), so, \(\mathrm{a}=6\)
Also passes through \((10,16)\) so, \(\frac{100}{36}-\frac{256}{\mathrm{~b}^2}=1\)
\(b^2=144\)
\(\therefore\) Equation \(\frac{\mathrm{x}^2}{36}-\frac{\mathrm{y}^2}{144}=1\)
Equation of normal is -
\(\frac{a^2 x}{x_1}+\frac{b^2}{y_1} y=a^2+b^2\)
\(\mathrm{x}_1=10, \mathrm{y}_1=16\)
\(\frac{36 x}{10}+\frac{144 y}{16}=144+36\)
\(\therefore \quad 2 x+5 y=100\)
(from solving)
120810
The equation of a common tangent to the curves, \(y^2=16 x\) and \(x y=-4\) is
1 \(x-y+4=0\)
2 \(x+y+4=0\)
3 \(x-2 y+16=0\)
4 \(2 x-y+2=0\)
Explanation:
A Given,
\(y=m x+\frac{4}{m}\) is always tangent to
\(y^2=16 x\)
If it is tangent to the \(x y=-4\)
\(x\left(m x+\frac{4}{m}\right)=-4\)
\(m^2 x^2+4 x=-4 m\)
\(m^2 x^2+4 x+4 m=0\)
For tangent \(\mathrm{D}=0\)
\(16-16 \mathrm{~m}^3=0 \Rightarrow \mathrm{m}=1 \text { put in equation (i) }\)
\(y=x+4\)
\(x-y+4=0\)
JEE Main 12.04.2019
Hyperbola
120811
Let \(P(3,3)\) be a point on the hyperbola, \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\). If the normal to it at \(P\) intersects the \(X\)-axis at \((9,0)\) and \(e\) is its eccentricity, then the ordered pair \(\left(a^2, \mathrm{e}^2\right)\) is equal to
1 \(\left(\frac{9}{2}, 3\right)\)
2 \(\left(\frac{3}{2}, 2\right)\)
3 \(\left(\frac{9}{2}, 2\right)\)
4 \((9,3)\)
Explanation:
A Given,
According to question,
\(\mathrm{p}(3,3) \text { lies on } \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
So, \(\quad \frac{9}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=1\)
Normal at \((3,3)\) is \(\rightarrow\)
\(y-3=-\frac{a^2}{b^2}(x-3)\)
Passes through \((9,0) \rightarrow\)
\(\mathrm{b}^2=2 \mathrm{a}^2\)
So,
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=3\)
Also,
\(\mathrm{a}^2=\frac{9}{2} \quad\) \{from equation (i) \& equation (ii) \(\}\)
Thus, \(\quad\left(\mathrm{a}^2, \mathrm{e}^2\right)=\left(\frac{9}{2}, 3\right)\)
JEE Main 04.09.2020
Hyperbola
120812
If the line \(y=m x+c\) is common tangent to the hyperbola \(\frac{x^2}{100}-\frac{y^2}{64}=1\) and the circle \(x^2+y^2=\) 36 , then which one of the following is true?
1 \(\mathrm{c}^2=369\)
2 \(5 \mathrm{~m}=4\)
3 \(4 \mathrm{c}^2=369\)
4 \(8 \mathrm{~m}+5=0\)
Explanation:
C Given,
Line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) is common tangent to the hyperbola
\(\frac{x^2}{100}-\frac{y^2}{64}=1\) and the circle \(x^2+y^2=36\)
We know that, for hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(c=\sqrt{a^2 m^2-b^2}\)
\(c=\sqrt{100 m^2-64}\)
For the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2, \mathrm{c}\) is given by \(\rightarrow\)
\(\mathrm{c}=\mathrm{a} \sqrt{1+\mathrm{m}^2}\)
\(\mathrm{c}=6 \sqrt{1+\mathrm{m}^2}\)
So, from equation (i) and equation (ii), we get -
\(\sqrt{100 \mathrm{~m}^2-64}=6 \sqrt{1+\mathrm{m}^2}\)
Square both side,
\(100 \mathrm{~m}^2-64=36\left(1+\mathrm{m}^2\right)\)
\(64 \mathrm{~m}^2=100\)
\(\mathrm{~m}= \pm \frac{10}{8}\)
From equation (ii) \(\rightarrow\)
\(c=6 \sqrt{1+\frac{100}{64}}\)
\(c=6 \sqrt{\frac{64+100}{64}}\)
\(c=6\left(\frac{1}{8} \sqrt{164}\right)\)
Taking square both side -
\(\mathrm{c}^2=36 \times\left(\frac{164}{64}\right)\)
\(\mathrm{c}^2=\frac{369}{4}\)
\(4 \mathrm{c}^2=369\)
JEE Main 05.09.2020
Hyperbola
120813
If a hyperbola passes through the point \(P(10\), \(16)\) and it has vertices at \(( \pm 6,0)\) then the equation of the normal to it at \(P\) is
1 \(3 x+4 y=94\)
2 \(x+2 y=42\)
3 \(2 x+5 y=100\)
4 \(x+3 y=58\)
Explanation:
C Given,
Let equation of hyperbola is -
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Given, its vertex at \(( \pm 6,0)\), so, \(\mathrm{a}=6\)
Also passes through \((10,16)\) so, \(\frac{100}{36}-\frac{256}{\mathrm{~b}^2}=1\)
\(b^2=144\)
\(\therefore\) Equation \(\frac{\mathrm{x}^2}{36}-\frac{\mathrm{y}^2}{144}=1\)
Equation of normal is -
\(\frac{a^2 x}{x_1}+\frac{b^2}{y_1} y=a^2+b^2\)
\(\mathrm{x}_1=10, \mathrm{y}_1=16\)
\(\frac{36 x}{10}+\frac{144 y}{16}=144+36\)
\(\therefore \quad 2 x+5 y=100\)
(from solving)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Hyperbola
120810
The equation of a common tangent to the curves, \(y^2=16 x\) and \(x y=-4\) is
1 \(x-y+4=0\)
2 \(x+y+4=0\)
3 \(x-2 y+16=0\)
4 \(2 x-y+2=0\)
Explanation:
A Given,
\(y=m x+\frac{4}{m}\) is always tangent to
\(y^2=16 x\)
If it is tangent to the \(x y=-4\)
\(x\left(m x+\frac{4}{m}\right)=-4\)
\(m^2 x^2+4 x=-4 m\)
\(m^2 x^2+4 x+4 m=0\)
For tangent \(\mathrm{D}=0\)
\(16-16 \mathrm{~m}^3=0 \Rightarrow \mathrm{m}=1 \text { put in equation (i) }\)
\(y=x+4\)
\(x-y+4=0\)
JEE Main 12.04.2019
Hyperbola
120811
Let \(P(3,3)\) be a point on the hyperbola, \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\). If the normal to it at \(P\) intersects the \(X\)-axis at \((9,0)\) and \(e\) is its eccentricity, then the ordered pair \(\left(a^2, \mathrm{e}^2\right)\) is equal to
1 \(\left(\frac{9}{2}, 3\right)\)
2 \(\left(\frac{3}{2}, 2\right)\)
3 \(\left(\frac{9}{2}, 2\right)\)
4 \((9,3)\)
Explanation:
A Given,
According to question,
\(\mathrm{p}(3,3) \text { lies on } \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
So, \(\quad \frac{9}{\mathrm{a}^2}-\frac{9}{\mathrm{~b}^2}=1\)
Normal at \((3,3)\) is \(\rightarrow\)
\(y-3=-\frac{a^2}{b^2}(x-3)\)
Passes through \((9,0) \rightarrow\)
\(\mathrm{b}^2=2 \mathrm{a}^2\)
So,
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=3\)
Also,
\(\mathrm{a}^2=\frac{9}{2} \quad\) \{from equation (i) \& equation (ii) \(\}\)
Thus, \(\quad\left(\mathrm{a}^2, \mathrm{e}^2\right)=\left(\frac{9}{2}, 3\right)\)
JEE Main 04.09.2020
Hyperbola
120812
If the line \(y=m x+c\) is common tangent to the hyperbola \(\frac{x^2}{100}-\frac{y^2}{64}=1\) and the circle \(x^2+y^2=\) 36 , then which one of the following is true?
1 \(\mathrm{c}^2=369\)
2 \(5 \mathrm{~m}=4\)
3 \(4 \mathrm{c}^2=369\)
4 \(8 \mathrm{~m}+5=0\)
Explanation:
C Given,
Line \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) is common tangent to the hyperbola
\(\frac{x^2}{100}-\frac{y^2}{64}=1\) and the circle \(x^2+y^2=36\)
We know that, for hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(c=\sqrt{a^2 m^2-b^2}\)
\(c=\sqrt{100 m^2-64}\)
For the circle \(\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2, \mathrm{c}\) is given by \(\rightarrow\)
\(\mathrm{c}=\mathrm{a} \sqrt{1+\mathrm{m}^2}\)
\(\mathrm{c}=6 \sqrt{1+\mathrm{m}^2}\)
So, from equation (i) and equation (ii), we get -
\(\sqrt{100 \mathrm{~m}^2-64}=6 \sqrt{1+\mathrm{m}^2}\)
Square both side,
\(100 \mathrm{~m}^2-64=36\left(1+\mathrm{m}^2\right)\)
\(64 \mathrm{~m}^2=100\)
\(\mathrm{~m}= \pm \frac{10}{8}\)
From equation (ii) \(\rightarrow\)
\(c=6 \sqrt{1+\frac{100}{64}}\)
\(c=6 \sqrt{\frac{64+100}{64}}\)
\(c=6\left(\frac{1}{8} \sqrt{164}\right)\)
Taking square both side -
\(\mathrm{c}^2=36 \times\left(\frac{164}{64}\right)\)
\(\mathrm{c}^2=\frac{369}{4}\)
\(4 \mathrm{c}^2=369\)
JEE Main 05.09.2020
Hyperbola
120813
If a hyperbola passes through the point \(P(10\), \(16)\) and it has vertices at \(( \pm 6,0)\) then the equation of the normal to it at \(P\) is
1 \(3 x+4 y=94\)
2 \(x+2 y=42\)
3 \(2 x+5 y=100\)
4 \(x+3 y=58\)
Explanation:
C Given,
Let equation of hyperbola is -
\(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\)
Given, its vertex at \(( \pm 6,0)\), so, \(\mathrm{a}=6\)
Also passes through \((10,16)\) so, \(\frac{100}{36}-\frac{256}{\mathrm{~b}^2}=1\)
\(b^2=144\)
\(\therefore\) Equation \(\frac{\mathrm{x}^2}{36}-\frac{\mathrm{y}^2}{144}=1\)
Equation of normal is -
\(\frac{a^2 x}{x_1}+\frac{b^2}{y_1} y=a^2+b^2\)
\(\mathrm{x}_1=10, \mathrm{y}_1=16\)
\(\frac{36 x}{10}+\frac{144 y}{16}=144+36\)
\(\therefore \quad 2 x+5 y=100\)
(from solving)