NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120780
The slope of common tangents of hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\) and \(\frac{y^2}{9}-\frac{x^2}{16}=1\) is
1 \(2,-2\)
2 \(1,-1\)
3 1,2
4 \(-1,-2\)
Explanation:
B Given,
Two hyperbolas are \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
and \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
Equation of tangent to (i) and having slope \(\mathrm{m}\) is
\(y=m x \pm \sqrt{9 m^2-16}\)
From equation (ii) and (iii), we get.-
\(\Rightarrow 16\left(\mathrm{mx} \pm \sqrt{9 \mathrm{~m}^2-16}\right)-9 \mathrm{x}^2=144\)
\(\left(16 \mathrm{~m}^2-9\right) \mathrm{x}^2 \pm 32 \mathrm{~m} \sqrt{9 \mathrm{~m}^2-16}+\left(144 \mathrm{~m}^2-400\right)=0\)
For it to be tangent, we must have, \(\mathrm{D}=0\)
\(\therefore\left(32 \mathrm{~m} \sqrt{9 \mathrm{~m}^2-16}\right)^2=4\left(16 \mathrm{~m}^2-9\right)\left(144 \mathrm{~m}^2-400\right)\)
\(\mathrm{m}^2=1\)
\(\mathrm{~m}= \pm 1 \text { or } 1,-1\)
UPSEE-2010
Hyperbola
120781
The equation of the tangent parallel to \(y-x+5\) \(=0\) drawn to \(\frac{x^2}{3}-\frac{y^2}{2}=1\) is
1 \(x-y+1=0\)
2 \(x-y+2=0\)
3 \(x+y-1=0\)
4 \(x+y+2=0\)
Explanation:
A The equation of any straight line, parallel to the given straight line \((\mathrm{y}=\mathrm{x}-5)\) will be \(\mathrm{y}=\mathrm{x}+\mathrm{c}[\because \mathrm{m}=1]\)
and equation of hyperbola \(\frac{x^2}{3}-\frac{y^2}{2}=1\)
\(a^2=3, b^2=2\)
Putting these values in \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\) \(c^2=3(1)^2-2=1 \therefore c= \pm 1\)
Hence, the equation of the required tangent will be \(\mathrm{y}=\mathrm{x} \pm 1 \Rightarrow \mathrm{x}-\mathrm{y}-1=0\) or \(\mathrm{x}-\mathrm{y}+1=0\)
JCECE-2008
Hyperbola
120782
The equation of the normal to the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) at the point \((8,3 \sqrt{3})\) is
1 \(\sqrt{3} x+2 y=25\)
2 \(2 x+\sqrt{3} y=25\)
3 \(y+2 x=25\)
4 \(x+y=25\)
Explanation:
B The equation of the normal to the hyperbola at the point \((8,3 \sqrt{3})\) is given as \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
Equation of the normal to the parabola-
\(\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2\)
\(\Rightarrow a^2=16, b^2=9\)
\(\Rightarrow x_1=8 \quad y_1=3 \sqrt{3}\)
\(\text { Then, } \frac{16 x}{8}+\frac{9 y}{3 \sqrt{3}}=16+9\)
\(\Rightarrow 2 x+\sqrt{3} y=25\)
AMU-2002
Hyperbola
120783
The equation of the normal to the hyperbola \(x^2-16 y^2-2 x-64 y-72=0\) at the point \((-4,-3)\) is
1 \(5 \mathrm{x}+16 \mathrm{y}+79=0\)
2 \(16 x+5 y+97=0\)
3 \(16 \mathrm{x}+5 \mathrm{y}+79=0\)
4 \(5 \mathrm{x}+16 \mathrm{y}+97=0\)
Explanation:
C Given,
\(x^2-16 y^2-2 x-64 y-72=0\)
Differentiating w.r.t. \(\mathrm{x}\),
\(\Rightarrow \quad 2 x-32 y \frac{d y}{d x}-2-64 \frac{d y}{d x}=0\)
\(\text { At point }(-4,-3)\)
\(\quad 2(-4)-32(-3) \frac{d y}{d x}-2-64 \frac{d y}{d x}=0\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{10}{32}=\frac{5}{16}=m \quad \text { (slope) }\)
At point \((-4,-3)\)
\(2(-4)-32(-3) \frac{\mathrm{dy}}{\mathrm{dx}}-2-64 \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
\(\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{10}{32}=\frac{5}{16}=\mathrm{m} \quad \text { (slope) }\)
Slope of normal \(=\frac{-1}{\mathrm{~m}}=\frac{-16}{5}\)
Equation of normal \(y-(-3)=\frac{-16}{5}(x-(-4))\)
\(\Rightarrow \quad y+3=\frac{-16}{5}(x+4)\)
\(16 x+5 y+79=0\)
120780
The slope of common tangents of hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\) and \(\frac{y^2}{9}-\frac{x^2}{16}=1\) is
1 \(2,-2\)
2 \(1,-1\)
3 1,2
4 \(-1,-2\)
Explanation:
B Given,
Two hyperbolas are \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
and \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
Equation of tangent to (i) and having slope \(\mathrm{m}\) is
\(y=m x \pm \sqrt{9 m^2-16}\)
From equation (ii) and (iii), we get.-
\(\Rightarrow 16\left(\mathrm{mx} \pm \sqrt{9 \mathrm{~m}^2-16}\right)-9 \mathrm{x}^2=144\)
\(\left(16 \mathrm{~m}^2-9\right) \mathrm{x}^2 \pm 32 \mathrm{~m} \sqrt{9 \mathrm{~m}^2-16}+\left(144 \mathrm{~m}^2-400\right)=0\)
For it to be tangent, we must have, \(\mathrm{D}=0\)
\(\therefore\left(32 \mathrm{~m} \sqrt{9 \mathrm{~m}^2-16}\right)^2=4\left(16 \mathrm{~m}^2-9\right)\left(144 \mathrm{~m}^2-400\right)\)
\(\mathrm{m}^2=1\)
\(\mathrm{~m}= \pm 1 \text { or } 1,-1\)
UPSEE-2010
Hyperbola
120781
The equation of the tangent parallel to \(y-x+5\) \(=0\) drawn to \(\frac{x^2}{3}-\frac{y^2}{2}=1\) is
1 \(x-y+1=0\)
2 \(x-y+2=0\)
3 \(x+y-1=0\)
4 \(x+y+2=0\)
Explanation:
A The equation of any straight line, parallel to the given straight line \((\mathrm{y}=\mathrm{x}-5)\) will be \(\mathrm{y}=\mathrm{x}+\mathrm{c}[\because \mathrm{m}=1]\)
and equation of hyperbola \(\frac{x^2}{3}-\frac{y^2}{2}=1\)
\(a^2=3, b^2=2\)
Putting these values in \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\) \(c^2=3(1)^2-2=1 \therefore c= \pm 1\)
Hence, the equation of the required tangent will be \(\mathrm{y}=\mathrm{x} \pm 1 \Rightarrow \mathrm{x}-\mathrm{y}-1=0\) or \(\mathrm{x}-\mathrm{y}+1=0\)
JCECE-2008
Hyperbola
120782
The equation of the normal to the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) at the point \((8,3 \sqrt{3})\) is
1 \(\sqrt{3} x+2 y=25\)
2 \(2 x+\sqrt{3} y=25\)
3 \(y+2 x=25\)
4 \(x+y=25\)
Explanation:
B The equation of the normal to the hyperbola at the point \((8,3 \sqrt{3})\) is given as \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
Equation of the normal to the parabola-
\(\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2\)
\(\Rightarrow a^2=16, b^2=9\)
\(\Rightarrow x_1=8 \quad y_1=3 \sqrt{3}\)
\(\text { Then, } \frac{16 x}{8}+\frac{9 y}{3 \sqrt{3}}=16+9\)
\(\Rightarrow 2 x+\sqrt{3} y=25\)
AMU-2002
Hyperbola
120783
The equation of the normal to the hyperbola \(x^2-16 y^2-2 x-64 y-72=0\) at the point \((-4,-3)\) is
1 \(5 \mathrm{x}+16 \mathrm{y}+79=0\)
2 \(16 x+5 y+97=0\)
3 \(16 \mathrm{x}+5 \mathrm{y}+79=0\)
4 \(5 \mathrm{x}+16 \mathrm{y}+97=0\)
Explanation:
C Given,
\(x^2-16 y^2-2 x-64 y-72=0\)
Differentiating w.r.t. \(\mathrm{x}\),
\(\Rightarrow \quad 2 x-32 y \frac{d y}{d x}-2-64 \frac{d y}{d x}=0\)
\(\text { At point }(-4,-3)\)
\(\quad 2(-4)-32(-3) \frac{d y}{d x}-2-64 \frac{d y}{d x}=0\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{10}{32}=\frac{5}{16}=m \quad \text { (slope) }\)
At point \((-4,-3)\)
\(2(-4)-32(-3) \frac{\mathrm{dy}}{\mathrm{dx}}-2-64 \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
\(\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{10}{32}=\frac{5}{16}=\mathrm{m} \quad \text { (slope) }\)
Slope of normal \(=\frac{-1}{\mathrm{~m}}=\frac{-16}{5}\)
Equation of normal \(y-(-3)=\frac{-16}{5}(x-(-4))\)
\(\Rightarrow \quad y+3=\frac{-16}{5}(x+4)\)
\(16 x+5 y+79=0\)
120780
The slope of common tangents of hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\) and \(\frac{y^2}{9}-\frac{x^2}{16}=1\) is
1 \(2,-2\)
2 \(1,-1\)
3 1,2
4 \(-1,-2\)
Explanation:
B Given,
Two hyperbolas are \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
and \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
Equation of tangent to (i) and having slope \(\mathrm{m}\) is
\(y=m x \pm \sqrt{9 m^2-16}\)
From equation (ii) and (iii), we get.-
\(\Rightarrow 16\left(\mathrm{mx} \pm \sqrt{9 \mathrm{~m}^2-16}\right)-9 \mathrm{x}^2=144\)
\(\left(16 \mathrm{~m}^2-9\right) \mathrm{x}^2 \pm 32 \mathrm{~m} \sqrt{9 \mathrm{~m}^2-16}+\left(144 \mathrm{~m}^2-400\right)=0\)
For it to be tangent, we must have, \(\mathrm{D}=0\)
\(\therefore\left(32 \mathrm{~m} \sqrt{9 \mathrm{~m}^2-16}\right)^2=4\left(16 \mathrm{~m}^2-9\right)\left(144 \mathrm{~m}^2-400\right)\)
\(\mathrm{m}^2=1\)
\(\mathrm{~m}= \pm 1 \text { or } 1,-1\)
UPSEE-2010
Hyperbola
120781
The equation of the tangent parallel to \(y-x+5\) \(=0\) drawn to \(\frac{x^2}{3}-\frac{y^2}{2}=1\) is
1 \(x-y+1=0\)
2 \(x-y+2=0\)
3 \(x+y-1=0\)
4 \(x+y+2=0\)
Explanation:
A The equation of any straight line, parallel to the given straight line \((\mathrm{y}=\mathrm{x}-5)\) will be \(\mathrm{y}=\mathrm{x}+\mathrm{c}[\because \mathrm{m}=1]\)
and equation of hyperbola \(\frac{x^2}{3}-\frac{y^2}{2}=1\)
\(a^2=3, b^2=2\)
Putting these values in \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\) \(c^2=3(1)^2-2=1 \therefore c= \pm 1\)
Hence, the equation of the required tangent will be \(\mathrm{y}=\mathrm{x} \pm 1 \Rightarrow \mathrm{x}-\mathrm{y}-1=0\) or \(\mathrm{x}-\mathrm{y}+1=0\)
JCECE-2008
Hyperbola
120782
The equation of the normal to the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) at the point \((8,3 \sqrt{3})\) is
1 \(\sqrt{3} x+2 y=25\)
2 \(2 x+\sqrt{3} y=25\)
3 \(y+2 x=25\)
4 \(x+y=25\)
Explanation:
B The equation of the normal to the hyperbola at the point \((8,3 \sqrt{3})\) is given as \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
Equation of the normal to the parabola-
\(\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2\)
\(\Rightarrow a^2=16, b^2=9\)
\(\Rightarrow x_1=8 \quad y_1=3 \sqrt{3}\)
\(\text { Then, } \frac{16 x}{8}+\frac{9 y}{3 \sqrt{3}}=16+9\)
\(\Rightarrow 2 x+\sqrt{3} y=25\)
AMU-2002
Hyperbola
120783
The equation of the normal to the hyperbola \(x^2-16 y^2-2 x-64 y-72=0\) at the point \((-4,-3)\) is
1 \(5 \mathrm{x}+16 \mathrm{y}+79=0\)
2 \(16 x+5 y+97=0\)
3 \(16 \mathrm{x}+5 \mathrm{y}+79=0\)
4 \(5 \mathrm{x}+16 \mathrm{y}+97=0\)
Explanation:
C Given,
\(x^2-16 y^2-2 x-64 y-72=0\)
Differentiating w.r.t. \(\mathrm{x}\),
\(\Rightarrow \quad 2 x-32 y \frac{d y}{d x}-2-64 \frac{d y}{d x}=0\)
\(\text { At point }(-4,-3)\)
\(\quad 2(-4)-32(-3) \frac{d y}{d x}-2-64 \frac{d y}{d x}=0\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{10}{32}=\frac{5}{16}=m \quad \text { (slope) }\)
At point \((-4,-3)\)
\(2(-4)-32(-3) \frac{\mathrm{dy}}{\mathrm{dx}}-2-64 \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
\(\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{10}{32}=\frac{5}{16}=\mathrm{m} \quad \text { (slope) }\)
Slope of normal \(=\frac{-1}{\mathrm{~m}}=\frac{-16}{5}\)
Equation of normal \(y-(-3)=\frac{-16}{5}(x-(-4))\)
\(\Rightarrow \quad y+3=\frac{-16}{5}(x+4)\)
\(16 x+5 y+79=0\)
120780
The slope of common tangents of hyperbola \(\frac{x^2}{9}-\frac{y^2}{16}=1\) and \(\frac{y^2}{9}-\frac{x^2}{16}=1\) is
1 \(2,-2\)
2 \(1,-1\)
3 1,2
4 \(-1,-2\)
Explanation:
B Given,
Two hyperbolas are \(\frac{x^2}{9}-\frac{y^2}{16}=1\)
and \(\frac{y^2}{9}-\frac{x^2}{16}=1\)
Equation of tangent to (i) and having slope \(\mathrm{m}\) is
\(y=m x \pm \sqrt{9 m^2-16}\)
From equation (ii) and (iii), we get.-
\(\Rightarrow 16\left(\mathrm{mx} \pm \sqrt{9 \mathrm{~m}^2-16}\right)-9 \mathrm{x}^2=144\)
\(\left(16 \mathrm{~m}^2-9\right) \mathrm{x}^2 \pm 32 \mathrm{~m} \sqrt{9 \mathrm{~m}^2-16}+\left(144 \mathrm{~m}^2-400\right)=0\)
For it to be tangent, we must have, \(\mathrm{D}=0\)
\(\therefore\left(32 \mathrm{~m} \sqrt{9 \mathrm{~m}^2-16}\right)^2=4\left(16 \mathrm{~m}^2-9\right)\left(144 \mathrm{~m}^2-400\right)\)
\(\mathrm{m}^2=1\)
\(\mathrm{~m}= \pm 1 \text { or } 1,-1\)
UPSEE-2010
Hyperbola
120781
The equation of the tangent parallel to \(y-x+5\) \(=0\) drawn to \(\frac{x^2}{3}-\frac{y^2}{2}=1\) is
1 \(x-y+1=0\)
2 \(x-y+2=0\)
3 \(x+y-1=0\)
4 \(x+y+2=0\)
Explanation:
A The equation of any straight line, parallel to the given straight line \((\mathrm{y}=\mathrm{x}-5)\) will be \(\mathrm{y}=\mathrm{x}+\mathrm{c}[\because \mathrm{m}=1]\)
and equation of hyperbola \(\frac{x^2}{3}-\frac{y^2}{2}=1\)
\(a^2=3, b^2=2\)
Putting these values in \(\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2\) \(c^2=3(1)^2-2=1 \therefore c= \pm 1\)
Hence, the equation of the required tangent will be \(\mathrm{y}=\mathrm{x} \pm 1 \Rightarrow \mathrm{x}-\mathrm{y}-1=0\) or \(\mathrm{x}-\mathrm{y}+1=0\)
JCECE-2008
Hyperbola
120782
The equation of the normal to the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) at the point \((8,3 \sqrt{3})\) is
1 \(\sqrt{3} x+2 y=25\)
2 \(2 x+\sqrt{3} y=25\)
3 \(y+2 x=25\)
4 \(x+y=25\)
Explanation:
B The equation of the normal to the hyperbola at the point \((8,3 \sqrt{3})\) is given as \(\frac{x^2}{16}-\frac{y^2}{9}=1\)
Equation of the normal to the parabola-
\(\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2\)
\(\Rightarrow a^2=16, b^2=9\)
\(\Rightarrow x_1=8 \quad y_1=3 \sqrt{3}\)
\(\text { Then, } \frac{16 x}{8}+\frac{9 y}{3 \sqrt{3}}=16+9\)
\(\Rightarrow 2 x+\sqrt{3} y=25\)
AMU-2002
Hyperbola
120783
The equation of the normal to the hyperbola \(x^2-16 y^2-2 x-64 y-72=0\) at the point \((-4,-3)\) is
1 \(5 \mathrm{x}+16 \mathrm{y}+79=0\)
2 \(16 x+5 y+97=0\)
3 \(16 \mathrm{x}+5 \mathrm{y}+79=0\)
4 \(5 \mathrm{x}+16 \mathrm{y}+97=0\)
Explanation:
C Given,
\(x^2-16 y^2-2 x-64 y-72=0\)
Differentiating w.r.t. \(\mathrm{x}\),
\(\Rightarrow \quad 2 x-32 y \frac{d y}{d x}-2-64 \frac{d y}{d x}=0\)
\(\text { At point }(-4,-3)\)
\(\quad 2(-4)-32(-3) \frac{d y}{d x}-2-64 \frac{d y}{d x}=0\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{10}{32}=\frac{5}{16}=m \quad \text { (slope) }\)
At point \((-4,-3)\)
\(2(-4)-32(-3) \frac{\mathrm{dy}}{\mathrm{dx}}-2-64 \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
\(\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{10}{32}=\frac{5}{16}=\mathrm{m} \quad \text { (slope) }\)
Slope of normal \(=\frac{-1}{\mathrm{~m}}=\frac{-16}{5}\)
Equation of normal \(y-(-3)=\frac{-16}{5}(x-(-4))\)
\(\Rightarrow \quad y+3=\frac{-16}{5}(x+4)\)
\(16 x+5 y+79=0\)
