88367
If \(\mathbf{m}\) is the slope of one of the lines represented by \(a x^{2}+2 h x y+b y^{2}=0\), then \((h+b m)^{2}=\)
1 \((a-b)^{2}\)
2 \((a+b)^{2}\)
3 \(\mathrm{h}^{2}-\mathrm{ab}\)
4 \(\mathrm{h}^{2}+\mathrm{ab}\)
Explanation:
(C) : Given, \(a x^{2}+2 h x y+b y^{2}=0\) is an homogeneous equation which shows pair of straight line and both equation are passing through the origin. Now, Slope of each equation is \(\mathrm{m}\) and \(\mathrm{m}_{1}\) \(\therefore\) The equation of straight line passing through origin \(\mathrm{y}=\mathrm{mx}\) \(y=m_{1} \mathrm{x}\) \(\therefore \quad \left(\mathrm{y}-\mathrm{mx}^{2}\right)\left(\mathrm{y}-\mathrm{m}_{1} \mathrm{x}\right)=0\) \(\mathrm{y}^{2}-\mathrm{m}_{1} \mathrm{xy}-\mathrm{mxy}^{2}+\mathrm{mm}_{1} \mathrm{x}^{2}=0\) \(\mathrm{~mm}_{1} \mathrm{x}^{2}-\left(\mathrm{m}_{1}+\mathrm{m}\right) \mathrm{x} . \mathrm{y}+\mathrm{y}^{2}=0 \tag{iii}\) Now, \(m+m_{1}=-\frac{2 h}{b}\) \(\mathrm{mm}_{1}=\mathrm{a} / \mathrm{b} \tag{v}\) From equation (iv) and (v) \(m_{1}=\left(\frac{-2 h}{b}-m\right)\) \(m\left(\frac{-2 h}{b}-m\right)=\frac{a}{b}\) \(\frac{-m}{b}(2 h+b m)=a / b\) \(-2 m h-m^{2} h=a\) \(-2 m h b-m^{2} b^{2}=a b\) \(h^{2}+2 m b h+m^{2} b^{2}=-a b+h^{2}\) \((h+m b)^{2}=h^{2}-a b\)
Karnataka CET-2010
Co-Ordinate system
88369
If the intercepts made on the line \(y=m x\) by lines \(y=2\) and \(y=6\) is less than 5 , then the range of values of \(m\) is :
1 \((-\infty,-4 / 3) \cup(4 / 3, \infty)\)
2 \((-4 / 3,4 / 3)\)
3 \((-3 / 4,4 / 3)\)
4 None of these
Explanation:
(A) : : Given, intercepts made on the line \(y=m x\) by the lines \(\mathrm{y}=2\) and \(\mathrm{y}=6\) is less than 5 \(\mathrm{y} \mathrm{mx} \rightarrow \mathrm{y}=2\) and \(\mathrm{y}=6\lt 5\) 'm' \(A\left(\frac{2}{m}, 2\right) B\left(\frac{6}{m}, 6\right)\lt 5\) \(\left(\frac{2}{m}-\frac{6}{m}\right)^{2}+(6-2)^{2}\lt 25\) \(\left(\frac{-4}{m}\right)^{2}+(4)^{2}\lt 25\) \(\frac{+16}{m^{2}}+16\lt 25\) \(\frac{16}{m^{2}}\lt 9\) \(\mathrm{m}^{2}>\frac{16}{4}\) \(\left(m-\frac{4}{3}\right)\left(m+\frac{4}{3}\right)>0\) \(\text { So, } \quad \mathrm{m} \in\left(\infty, \frac{-4}{3}\right) \cup\left(\frac{4}{3}, \infty\right)\)
AMU-2019
Co-Ordinate system
88370
What is the equation of the curve through the point \((1,1)\) and whose slope is \(\frac{2 a y}{x(y-a)}\) ?
(A) : From option (a), we have- \(y^{a} x^{2 a}=e^{y-1} \quad\)...........(i) Differentiating both sides w.r.t. to \(x\). \(\Rightarrow y^{\prime}=\frac{2 a y^{a} \cdot \frac{x^{2 a}}{x}}{e^{y-1}-a \frac{y^{a} \times 2 a}{y}}\) From (i) and (ii), \(y^{\prime}=\frac{y \cdot 2 a e^{y-1}}{x\left(y e^{y-1}-a e^{y-1}\right)}=\frac{2 a y}{x(y-a)}\) which, is the required equation.
SCRA-2012
Co-Ordinate system
88371
If the lines joining the origin to the intersection of the line \(y=m x+2\) and the circle \(x^{2}+y^{2}=1\) are at right angles, then
1 \(\mathrm{m}=\sqrt{3}\)
2 \(\mathrm{m}= \pm \sqrt{7}\)
3 \(\mathrm{m}=\sqrt{1}\)
4 \(\mathrm{m}=\sqrt{5}\)
Explanation:
(B) : Joint equation of the lines joining the origin and the point of intersection of the line \(\mathrm{y}=\mathrm{mx}+2\) and the curve \(x^{2}+y^{2}=1\) \(x^{2}+y^{2}=\left(\frac{y-m x}{2}\right)^{2}\) \(4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{y}^{2}+\mathrm{m}^{2} \mathrm{x}^{2}-2 \mathrm{ymx}\) \(x^{2}\left(4-m^{2}\right)+2 m x y+3 y^{2}=0\) Since these line are at right angles, So, \(\quad \mathrm{C}_{\mathrm{x}}+\mathrm{C}_{\mathrm{y}}=0\) \(4-m^{2}+3=0\) \(m^{2}=7\) \(m= \pm \sqrt{7}\)
88367
If \(\mathbf{m}\) is the slope of one of the lines represented by \(a x^{2}+2 h x y+b y^{2}=0\), then \((h+b m)^{2}=\)
1 \((a-b)^{2}\)
2 \((a+b)^{2}\)
3 \(\mathrm{h}^{2}-\mathrm{ab}\)
4 \(\mathrm{h}^{2}+\mathrm{ab}\)
Explanation:
(C) : Given, \(a x^{2}+2 h x y+b y^{2}=0\) is an homogeneous equation which shows pair of straight line and both equation are passing through the origin. Now, Slope of each equation is \(\mathrm{m}\) and \(\mathrm{m}_{1}\) \(\therefore\) The equation of straight line passing through origin \(\mathrm{y}=\mathrm{mx}\) \(y=m_{1} \mathrm{x}\) \(\therefore \quad \left(\mathrm{y}-\mathrm{mx}^{2}\right)\left(\mathrm{y}-\mathrm{m}_{1} \mathrm{x}\right)=0\) \(\mathrm{y}^{2}-\mathrm{m}_{1} \mathrm{xy}-\mathrm{mxy}^{2}+\mathrm{mm}_{1} \mathrm{x}^{2}=0\) \(\mathrm{~mm}_{1} \mathrm{x}^{2}-\left(\mathrm{m}_{1}+\mathrm{m}\right) \mathrm{x} . \mathrm{y}+\mathrm{y}^{2}=0 \tag{iii}\) Now, \(m+m_{1}=-\frac{2 h}{b}\) \(\mathrm{mm}_{1}=\mathrm{a} / \mathrm{b} \tag{v}\) From equation (iv) and (v) \(m_{1}=\left(\frac{-2 h}{b}-m\right)\) \(m\left(\frac{-2 h}{b}-m\right)=\frac{a}{b}\) \(\frac{-m}{b}(2 h+b m)=a / b\) \(-2 m h-m^{2} h=a\) \(-2 m h b-m^{2} b^{2}=a b\) \(h^{2}+2 m b h+m^{2} b^{2}=-a b+h^{2}\) \((h+m b)^{2}=h^{2}-a b\)
Karnataka CET-2010
Co-Ordinate system
88369
If the intercepts made on the line \(y=m x\) by lines \(y=2\) and \(y=6\) is less than 5 , then the range of values of \(m\) is :
1 \((-\infty,-4 / 3) \cup(4 / 3, \infty)\)
2 \((-4 / 3,4 / 3)\)
3 \((-3 / 4,4 / 3)\)
4 None of these
Explanation:
(A) : : Given, intercepts made on the line \(y=m x\) by the lines \(\mathrm{y}=2\) and \(\mathrm{y}=6\) is less than 5 \(\mathrm{y} \mathrm{mx} \rightarrow \mathrm{y}=2\) and \(\mathrm{y}=6\lt 5\) 'm' \(A\left(\frac{2}{m}, 2\right) B\left(\frac{6}{m}, 6\right)\lt 5\) \(\left(\frac{2}{m}-\frac{6}{m}\right)^{2}+(6-2)^{2}\lt 25\) \(\left(\frac{-4}{m}\right)^{2}+(4)^{2}\lt 25\) \(\frac{+16}{m^{2}}+16\lt 25\) \(\frac{16}{m^{2}}\lt 9\) \(\mathrm{m}^{2}>\frac{16}{4}\) \(\left(m-\frac{4}{3}\right)\left(m+\frac{4}{3}\right)>0\) \(\text { So, } \quad \mathrm{m} \in\left(\infty, \frac{-4}{3}\right) \cup\left(\frac{4}{3}, \infty\right)\)
AMU-2019
Co-Ordinate system
88370
What is the equation of the curve through the point \((1,1)\) and whose slope is \(\frac{2 a y}{x(y-a)}\) ?
(A) : From option (a), we have- \(y^{a} x^{2 a}=e^{y-1} \quad\)...........(i) Differentiating both sides w.r.t. to \(x\). \(\Rightarrow y^{\prime}=\frac{2 a y^{a} \cdot \frac{x^{2 a}}{x}}{e^{y-1}-a \frac{y^{a} \times 2 a}{y}}\) From (i) and (ii), \(y^{\prime}=\frac{y \cdot 2 a e^{y-1}}{x\left(y e^{y-1}-a e^{y-1}\right)}=\frac{2 a y}{x(y-a)}\) which, is the required equation.
SCRA-2012
Co-Ordinate system
88371
If the lines joining the origin to the intersection of the line \(y=m x+2\) and the circle \(x^{2}+y^{2}=1\) are at right angles, then
1 \(\mathrm{m}=\sqrt{3}\)
2 \(\mathrm{m}= \pm \sqrt{7}\)
3 \(\mathrm{m}=\sqrt{1}\)
4 \(\mathrm{m}=\sqrt{5}\)
Explanation:
(B) : Joint equation of the lines joining the origin and the point of intersection of the line \(\mathrm{y}=\mathrm{mx}+2\) and the curve \(x^{2}+y^{2}=1\) \(x^{2}+y^{2}=\left(\frac{y-m x}{2}\right)^{2}\) \(4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{y}^{2}+\mathrm{m}^{2} \mathrm{x}^{2}-2 \mathrm{ymx}\) \(x^{2}\left(4-m^{2}\right)+2 m x y+3 y^{2}=0\) Since these line are at right angles, So, \(\quad \mathrm{C}_{\mathrm{x}}+\mathrm{C}_{\mathrm{y}}=0\) \(4-m^{2}+3=0\) \(m^{2}=7\) \(m= \pm \sqrt{7}\)
88367
If \(\mathbf{m}\) is the slope of one of the lines represented by \(a x^{2}+2 h x y+b y^{2}=0\), then \((h+b m)^{2}=\)
1 \((a-b)^{2}\)
2 \((a+b)^{2}\)
3 \(\mathrm{h}^{2}-\mathrm{ab}\)
4 \(\mathrm{h}^{2}+\mathrm{ab}\)
Explanation:
(C) : Given, \(a x^{2}+2 h x y+b y^{2}=0\) is an homogeneous equation which shows pair of straight line and both equation are passing through the origin. Now, Slope of each equation is \(\mathrm{m}\) and \(\mathrm{m}_{1}\) \(\therefore\) The equation of straight line passing through origin \(\mathrm{y}=\mathrm{mx}\) \(y=m_{1} \mathrm{x}\) \(\therefore \quad \left(\mathrm{y}-\mathrm{mx}^{2}\right)\left(\mathrm{y}-\mathrm{m}_{1} \mathrm{x}\right)=0\) \(\mathrm{y}^{2}-\mathrm{m}_{1} \mathrm{xy}-\mathrm{mxy}^{2}+\mathrm{mm}_{1} \mathrm{x}^{2}=0\) \(\mathrm{~mm}_{1} \mathrm{x}^{2}-\left(\mathrm{m}_{1}+\mathrm{m}\right) \mathrm{x} . \mathrm{y}+\mathrm{y}^{2}=0 \tag{iii}\) Now, \(m+m_{1}=-\frac{2 h}{b}\) \(\mathrm{mm}_{1}=\mathrm{a} / \mathrm{b} \tag{v}\) From equation (iv) and (v) \(m_{1}=\left(\frac{-2 h}{b}-m\right)\) \(m\left(\frac{-2 h}{b}-m\right)=\frac{a}{b}\) \(\frac{-m}{b}(2 h+b m)=a / b\) \(-2 m h-m^{2} h=a\) \(-2 m h b-m^{2} b^{2}=a b\) \(h^{2}+2 m b h+m^{2} b^{2}=-a b+h^{2}\) \((h+m b)^{2}=h^{2}-a b\)
Karnataka CET-2010
Co-Ordinate system
88369
If the intercepts made on the line \(y=m x\) by lines \(y=2\) and \(y=6\) is less than 5 , then the range of values of \(m\) is :
1 \((-\infty,-4 / 3) \cup(4 / 3, \infty)\)
2 \((-4 / 3,4 / 3)\)
3 \((-3 / 4,4 / 3)\)
4 None of these
Explanation:
(A) : : Given, intercepts made on the line \(y=m x\) by the lines \(\mathrm{y}=2\) and \(\mathrm{y}=6\) is less than 5 \(\mathrm{y} \mathrm{mx} \rightarrow \mathrm{y}=2\) and \(\mathrm{y}=6\lt 5\) 'm' \(A\left(\frac{2}{m}, 2\right) B\left(\frac{6}{m}, 6\right)\lt 5\) \(\left(\frac{2}{m}-\frac{6}{m}\right)^{2}+(6-2)^{2}\lt 25\) \(\left(\frac{-4}{m}\right)^{2}+(4)^{2}\lt 25\) \(\frac{+16}{m^{2}}+16\lt 25\) \(\frac{16}{m^{2}}\lt 9\) \(\mathrm{m}^{2}>\frac{16}{4}\) \(\left(m-\frac{4}{3}\right)\left(m+\frac{4}{3}\right)>0\) \(\text { So, } \quad \mathrm{m} \in\left(\infty, \frac{-4}{3}\right) \cup\left(\frac{4}{3}, \infty\right)\)
AMU-2019
Co-Ordinate system
88370
What is the equation of the curve through the point \((1,1)\) and whose slope is \(\frac{2 a y}{x(y-a)}\) ?
(A) : From option (a), we have- \(y^{a} x^{2 a}=e^{y-1} \quad\)...........(i) Differentiating both sides w.r.t. to \(x\). \(\Rightarrow y^{\prime}=\frac{2 a y^{a} \cdot \frac{x^{2 a}}{x}}{e^{y-1}-a \frac{y^{a} \times 2 a}{y}}\) From (i) and (ii), \(y^{\prime}=\frac{y \cdot 2 a e^{y-1}}{x\left(y e^{y-1}-a e^{y-1}\right)}=\frac{2 a y}{x(y-a)}\) which, is the required equation.
SCRA-2012
Co-Ordinate system
88371
If the lines joining the origin to the intersection of the line \(y=m x+2\) and the circle \(x^{2}+y^{2}=1\) are at right angles, then
1 \(\mathrm{m}=\sqrt{3}\)
2 \(\mathrm{m}= \pm \sqrt{7}\)
3 \(\mathrm{m}=\sqrt{1}\)
4 \(\mathrm{m}=\sqrt{5}\)
Explanation:
(B) : Joint equation of the lines joining the origin and the point of intersection of the line \(\mathrm{y}=\mathrm{mx}+2\) and the curve \(x^{2}+y^{2}=1\) \(x^{2}+y^{2}=\left(\frac{y-m x}{2}\right)^{2}\) \(4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{y}^{2}+\mathrm{m}^{2} \mathrm{x}^{2}-2 \mathrm{ymx}\) \(x^{2}\left(4-m^{2}\right)+2 m x y+3 y^{2}=0\) Since these line are at right angles, So, \(\quad \mathrm{C}_{\mathrm{x}}+\mathrm{C}_{\mathrm{y}}=0\) \(4-m^{2}+3=0\) \(m^{2}=7\) \(m= \pm \sqrt{7}\)
88367
If \(\mathbf{m}\) is the slope of one of the lines represented by \(a x^{2}+2 h x y+b y^{2}=0\), then \((h+b m)^{2}=\)
1 \((a-b)^{2}\)
2 \((a+b)^{2}\)
3 \(\mathrm{h}^{2}-\mathrm{ab}\)
4 \(\mathrm{h}^{2}+\mathrm{ab}\)
Explanation:
(C) : Given, \(a x^{2}+2 h x y+b y^{2}=0\) is an homogeneous equation which shows pair of straight line and both equation are passing through the origin. Now, Slope of each equation is \(\mathrm{m}\) and \(\mathrm{m}_{1}\) \(\therefore\) The equation of straight line passing through origin \(\mathrm{y}=\mathrm{mx}\) \(y=m_{1} \mathrm{x}\) \(\therefore \quad \left(\mathrm{y}-\mathrm{mx}^{2}\right)\left(\mathrm{y}-\mathrm{m}_{1} \mathrm{x}\right)=0\) \(\mathrm{y}^{2}-\mathrm{m}_{1} \mathrm{xy}-\mathrm{mxy}^{2}+\mathrm{mm}_{1} \mathrm{x}^{2}=0\) \(\mathrm{~mm}_{1} \mathrm{x}^{2}-\left(\mathrm{m}_{1}+\mathrm{m}\right) \mathrm{x} . \mathrm{y}+\mathrm{y}^{2}=0 \tag{iii}\) Now, \(m+m_{1}=-\frac{2 h}{b}\) \(\mathrm{mm}_{1}=\mathrm{a} / \mathrm{b} \tag{v}\) From equation (iv) and (v) \(m_{1}=\left(\frac{-2 h}{b}-m\right)\) \(m\left(\frac{-2 h}{b}-m\right)=\frac{a}{b}\) \(\frac{-m}{b}(2 h+b m)=a / b\) \(-2 m h-m^{2} h=a\) \(-2 m h b-m^{2} b^{2}=a b\) \(h^{2}+2 m b h+m^{2} b^{2}=-a b+h^{2}\) \((h+m b)^{2}=h^{2}-a b\)
Karnataka CET-2010
Co-Ordinate system
88369
If the intercepts made on the line \(y=m x\) by lines \(y=2\) and \(y=6\) is less than 5 , then the range of values of \(m\) is :
1 \((-\infty,-4 / 3) \cup(4 / 3, \infty)\)
2 \((-4 / 3,4 / 3)\)
3 \((-3 / 4,4 / 3)\)
4 None of these
Explanation:
(A) : : Given, intercepts made on the line \(y=m x\) by the lines \(\mathrm{y}=2\) and \(\mathrm{y}=6\) is less than 5 \(\mathrm{y} \mathrm{mx} \rightarrow \mathrm{y}=2\) and \(\mathrm{y}=6\lt 5\) 'm' \(A\left(\frac{2}{m}, 2\right) B\left(\frac{6}{m}, 6\right)\lt 5\) \(\left(\frac{2}{m}-\frac{6}{m}\right)^{2}+(6-2)^{2}\lt 25\) \(\left(\frac{-4}{m}\right)^{2}+(4)^{2}\lt 25\) \(\frac{+16}{m^{2}}+16\lt 25\) \(\frac{16}{m^{2}}\lt 9\) \(\mathrm{m}^{2}>\frac{16}{4}\) \(\left(m-\frac{4}{3}\right)\left(m+\frac{4}{3}\right)>0\) \(\text { So, } \quad \mathrm{m} \in\left(\infty, \frac{-4}{3}\right) \cup\left(\frac{4}{3}, \infty\right)\)
AMU-2019
Co-Ordinate system
88370
What is the equation of the curve through the point \((1,1)\) and whose slope is \(\frac{2 a y}{x(y-a)}\) ?
(A) : From option (a), we have- \(y^{a} x^{2 a}=e^{y-1} \quad\)...........(i) Differentiating both sides w.r.t. to \(x\). \(\Rightarrow y^{\prime}=\frac{2 a y^{a} \cdot \frac{x^{2 a}}{x}}{e^{y-1}-a \frac{y^{a} \times 2 a}{y}}\) From (i) and (ii), \(y^{\prime}=\frac{y \cdot 2 a e^{y-1}}{x\left(y e^{y-1}-a e^{y-1}\right)}=\frac{2 a y}{x(y-a)}\) which, is the required equation.
SCRA-2012
Co-Ordinate system
88371
If the lines joining the origin to the intersection of the line \(y=m x+2\) and the circle \(x^{2}+y^{2}=1\) are at right angles, then
1 \(\mathrm{m}=\sqrt{3}\)
2 \(\mathrm{m}= \pm \sqrt{7}\)
3 \(\mathrm{m}=\sqrt{1}\)
4 \(\mathrm{m}=\sqrt{5}\)
Explanation:
(B) : Joint equation of the lines joining the origin and the point of intersection of the line \(\mathrm{y}=\mathrm{mx}+2\) and the curve \(x^{2}+y^{2}=1\) \(x^{2}+y^{2}=\left(\frac{y-m x}{2}\right)^{2}\) \(4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathrm{y}^{2}+\mathrm{m}^{2} \mathrm{x}^{2}-2 \mathrm{ymx}\) \(x^{2}\left(4-m^{2}\right)+2 m x y+3 y^{2}=0\) Since these line are at right angles, So, \(\quad \mathrm{C}_{\mathrm{x}}+\mathrm{C}_{\mathrm{y}}=0\) \(4-m^{2}+3=0\) \(m^{2}=7\) \(m= \pm \sqrt{7}\)
