Differentiation (Part-38) | S N De

 

Previously, we have discussed twenty  mathematical  problems and their solutions which are important for preparation for WBCS Maths optional papers. . You can find part 1 and part 2 of those discussions .  Here, in this article , we will discuss few more problems .

21. If $\,\, ax^2+2hxy+by^2+2gx+2fy+c=0\,\,$ represent pair of straight lines, show that the area of the parallelogram formed by them and the pair of parallel lines through the origin is  $\,\frac{c}{2\sqrt{h^2-ab}}\,$

Sol. Let OACB be the parallelogram. The equation of OA and OB in pair form is           

 $\,\, ax^2+2hxy+by^2+2gx+2fy+c=0.\,\,$       

If $\,\,p_1,p_2\,\,$ are perpendicular distance from $\,\,C(\alpha,\beta)\,\,$ on OA and OB respectively       

and $\,\,\angle{AOB}=\theta\,\,$ then the required area of the parallelogram =$2 \Delta  OAC $                   

But ,  $\,\,2 \Delta OAC=\frac{1}{2}.OA. p_1=\frac{1}{2}.BC. p_1 \\ [\text{since,} OA=BC]$

Now, from $\,\,\Delta BLC,\,\,$ we get,          

 $\,\,\sin{\theta}=\frac{CL}{BC}, \text{where,}\,\,CL \perp OL \\ \Rightarrow \sin{\theta}=\frac{p_2}{BC} \\ \Rightarrow BC=\frac{p_2}{\sin{\theta}} $  

So, $\,\, \Delta OAC=\frac{1}{2}. \frac{p_2}{\sin{\theta}}.p_1 \\ \Rightarrow 2 \Delta OAC=\frac{p_1p_2}{\sin{\theta}} \cdots \cdots (*) $                

Let $ \,\, ax^2+2hxy+by^2 =by^2+2hxy+ax^2 \\ \equiv b(y-m_1x)(y-m_2x) \\ \Rightarrow y^2+\frac{2h}{b}xy+\frac{a}{b}.x^2=(y-m_1x)(y-m_2x) \\  [\longrightarrow (1)] \\=y^2-(m_1+m_2)xy+m_1m_2x^2$

From (1), we get , $\,\, m_1+m_2=-\frac{2h}{b},\,\,m_1m_2=\frac{a}{b}.$

Now, let OA be represented by $\,\,y-m_1x=0$                 

and OB be represented by  $\,\, y-m_2x=0$      

Then, $\,\,p_1=\frac{\beta-m_1\alpha}{\sqrt{1+m_1^2}} \\ \text{and}\,\, p_2=\frac{\beta-m_2\alpha}{\sqrt{1+m_2^2}}\,\, \text{so that} \\ p_1p_2= \frac{(\beta-m_1\alpha)(\beta-m_2\alpha)}{\sqrt{1+m_1^2}\sqrt{1+m_2^2}} \\=\frac{\beta^2-(m_1+m_2)\alpha\beta +(m_1m_2)\alpha^2}{\sqrt{1+m_1^2+m_2^2+m_1^2m_2^2}}\\=\frac{\beta^2+\frac{2h}{b}.\alpha\beta+\frac{a}{b}.\alpha^2}{\sqrt{1+(m_1+m_2)^2-2m_1m_2+m_1^2m_2^2}}\\=\frac{\beta^2+\frac{2h}{b}.\alpha\beta+\frac{a}{b}.\alpha^2}{\sqrt{1+(-2h/b)^2-2.\frac{a}{b}+\frac{a^2}{b^2}}}\\=\frac{b\beta^2+2h\alpha\beta+a\alpha^2}{\sqrt{1+4h^2-2ab+a^2}}\\=\frac{b\beta^2+2h\alpha\beta+a\alpha^2}{\sqrt{(a-b)^2+4h^2}}\\ \text{and}\,\,\sin{\theta}=\frac{2\sqrt{h^2-ab}}{\sqrt{(a-b)^2+4h^2}} \\ \quad [\text{Using the fact,}\tan{\theta}=\frac{2\sqrt{h^2-ab}}{a+b}]$

Now, from the given condition we see          

 $\,\,a \alpha^2+2h \alpha\beta+b \beta^2+2g \alpha+2f \beta+c=0  \\ ~~~~~~~~~~~~~\cdots (2)\\ \text{and}\,\,\,g\alpha+f \beta+c=0 \cdots (3)$ 

Now, from (3), we see           

$\,\,g \alpha+f \beta=-c\\ \text{From (2),}\,\,a \alpha^2+2h \alpha\beta+b \beta^2 \\+2(g \alpha+f \beta)+c=0 \\ \Rightarrow a \alpha^2+2h \alpha\beta+b \beta^2 +2(-c)+c=0 \\ \Rightarrow a \alpha^2+2h \alpha\beta+b\beta^2=c $

So, from (*), we get             

 $\\ \Delta OAC=\frac{p_1p_2}{\sin{\theta}}\\=\frac{(a\alpha^2+2h \alpha\beta+b \beta^2)/B}{2 \sqrt{h^2-ab}/B}\\ \text{where,}\,\,B=\sqrt{(a-b)^2+4h^2}\\=\frac{a\alpha^2+2h \alpha\beta+b \beta^2}{2\sqrt{h^2-ab}}\\=\frac{c}{2\sqrt{h^2-ab}}$

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22. Show that the two lines represented by the equation $\,\, ay^4+bxy^3+cx^2y^2 +dx^3y+ex^4=0 \,\,$ will be at right angles, if $\,\,(b+d)(ad+be)+(a-e)^2(a+c+e)=0.$

Sol. Since two of the lines represented by the given equation are at right angles, we assume that $\\ay^4+bxy^3+cx^2y^2+dx^3y+ex^4 \\=(ay^2+qxy+rx^2)(y^2+pyx-x^2) \longrightarrow (1) \\=ay^4+(ap+q)xy^3+(-a+pq+r)x^2y^2 \\+(-q+pr)x^3y+(-r)x^4 $

Now, comparing the coefficients of like powers both sides of (1),                 

$\\ b=ap+q,\,\,c=-a+pq+r,\\d=-q+pr,\,\,e=-r $ 

Now, $\,\,b+d=ap+q-q+pr=p(a+r) \\ ad+be=a(-q+pr)+(ap+q)(-r)\\=-aq+apr-apr-qr=-q(a+r)\\ \therefore (b+d)(ad+be)=-pq(a+r)^2 \\ (a-e)^2(a+c+e) \\=(a+r)^2\{a-r+(-a)+pq+r\} \\=pq(a+r)^2 \\ \therefore (b+d)(ad+be)+(a-e)^2(a+c+e) \\=-pq(a+r)^2 +pq(a+r)^2\\=0 $

 

23. If each of the equation  $\,\,f_1(x,y)=ax^2+2hxy+by^2+2gx+2fy+c=0 $   and $\,\, f_2(x,y)=ax^2+2hxy+by^2-2gx-2fy+c=0\,\,$ represents a pair of straight lines , prove that the area of the parallelogram enclosed by them is $\,\,\frac{2c}{\sqrt{h^2-ab}}$

Sol. Let $\,\, AD: l_1x+m_1y+n_1=0, \\DC:l_2x+m_2y+n_2=0 \\ ax^2+2hxy+by^2+2gx+2fy+c \\=(l_1x+m_1y+n_1)(l_2x+m_2y+n_2)  \longrightarrow (1)$ 

So, comparing the like coefficients from both sides of (1), we get         

$\,\, 2g=l_1n_2+l_2n_1,\,\, 2f=m_1n_2+m_2n_1 \longrightarrow (2)\\$                

If we replace $\,n_1\,$ and $\,n_2\,$ by $\,-n_1\,$ and $\,-n_2\,$, we get from (2)              

$\,\, -2g=l_1n_2+l_2n_1,\,\, -2f=m_1n_2+m_2n_1 \\$ .              

 So, equations of BC which is parallel to AD, can be taken as                   

 $BC: l_1x+m_1y-n_1=0 \\ AB : l_2x+m_2y-n_2=0. $ 

Let  $\angle{ABC}=\theta.\,\,\\ \therefore p_1=\frac{n_1-(-n_1)}{\sqrt{l_1^2+m_1^2}}=\frac{2n_1}{\sqrt{l_1^2+m_1^2}} \\ \text{and}\,\, p_2=\frac{n_2-(-n_2)}{\sqrt{l_2^2+m_2^2}} =\frac{2n_2}{\sqrt{l_2^2+m_2^2}}\\ \text{Now,}\,\,(l_1^2+m_1^2)(l_2^2+m_2^2)\\= (l_1l_2)^2+(m_1m_2)^2+(l_1m_2)^2+(l_2m_1)^2\\=a^2+b^2+(2h)^2-2.(l_1l_2).(m_1m_2)\\=a^2+b^2-2ab+4h^2\\=(a-b)^2+4h^2 \\ \therefore \sqrt{\{(l_1^2+m_1^2)(l_2^2+m_2^2)\}}= \sqrt{(a-b)^2+4h^2 } \\ \therefore p_1p_2=\frac{4c}{\sqrt{(a-b)^2+4h^2 }}\,\, [\text{since,}\,\,n_1n_2=c] \\ \tan{\theta}=\left|\frac{(-\frac{l_1}{m_1})-(\frac{-l_2}{m_2})}{1+\frac{l_1l_2}{m_1m_2}}\right|=\left|\frac{l_1m_2-l_2m_1}{m_1m_2+l_1l_2}\right| \longrightarrow(3)\\ \text{Now,}\,\, (l_1m_2-l_2m_1)^2\\=(l_1m_2+l_2m_1)^2-4l_1l_2m_1m_2 \\= (2h)^2-4ab\\=4(h^2-ab) $

So, from (3), we get $\,\,\tan{\theta}=\frac{2 \sqrt{h^2-ab}}{a+b} $                  

and so $\,\, \sin{\theta}=\frac{2 \sqrt{h^2-ab}}{\sqrt{(a-b)^2+4h^2}} $

So, the area of the parallelogram is  (A*)                               

$= \frac{p_1p_2}{\sin{\theta}}=\frac{4c/L}{2 \sqrt{h^2-ab}/L}\\  \text{where,}\,\, L=\sqrt{(l_1^2+m_1^2)(l_2^2+m_2^2)} \\ \Rightarrow \text{A*}=\frac{2c}{\sqrt{h^2-ab}}$

24.  A point P moves on the plane which is fixed.The plane through P and perpendicular to OP meets the axes in A,B,C and parallel to $\,\,YZ,\,\,ZX,\,\,XY\,\,$ planes respectively intersect in Q. Prove that if the axes be   rectangular , the locus of Q is $\,\, \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}$

Sol. Let $\,\,(\alpha, \beta,\gamma)\,$ be the coordinates of P.                       

Then $\,\, \frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}=1 \longrightarrow(1).$

 The equation of the plane through P and perpendicular to OP is $ \,\, \alpha x +\beta y+\gamma z=\alpha^2+\beta^2+\gamma^2 .$

The coordinates of  A,B,C are  $\,\, \left(\sum {\frac{\alpha^2}{\alpha}},0,0\right),\left(0,\sum {\frac{\alpha^2}{\beta}},0\right),\left(0,0,\sum {\frac{\alpha^2}{\gamma}}\right)\,\,$ respectively. 

 If $\,\,(x',y',z')\,\,$ be the coordinates of Q, then $x'= \frac{\alpha^2}{\alpha};\,\,y'=\frac{\alpha^2}{\beta};\,\,z'=\frac{\alpha^2}{\gamma}; \\ \text{Now,}\,\, \frac{1}{x'^2}+\frac{1}{y'^2}+\frac{1}{z'^2}=\frac{1}{\sum{\alpha^2}},\,\,\text{and} \\ \frac{1}{ax'}+\frac{1}{by'}+\frac{1}{cz'}=\frac{1}{\sum{\alpha^2}} \,\,[\text{By (1)}] \\$

Hence the locus of Q is $\,\, \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}$

25. Show that if one of the lines given by the equation $\,\,ax^2+2hxy+by^2=0\,\,$ coincides with one of those given by $\,\,a'x^2+2h'xy+b'y^2=0\,\,$ then $\,\, (ab'-a'b)^2=4(ha'-h'a)(bh'-b'h)$

Sol. Let $\,\, y=mx\,\,$ be the common line . So, $ax^2+2hx(mx)+b(mx)^2=0  \\ \Rightarrow bm^2+2hm+a=0 \\   a'x^2+2h'x(mx)+b'(mx)^2=0  \\ \Rightarrow b'm^2+2h'm+a'=0 $  

Hence, by cross-multiplication we get,                  

 $\,\, \frac{m^2}{2(ha'-h'a)} =\frac{m}{ab'-a'b}=\frac{1}{2(h'b-hb')} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \longrightarrow(1)$ 

So, from the last two relations we get, 

$ m=\frac{ab'-a'b}{2(h'b-hb')} \longrightarrow(2)$ 

Also, from the first two relations we get, 

 $m=2\frac{ha'-h'a}{ab'-a'b} \longrightarrow (3)  $  

Hence,  from (2) and (3), we get the required result.

26. Show that $\,\,8x^2+10xy+3y^2+26x+16y+21=0\,\,$ represent a pair of straight lines and find the point of intersections and angles between them.

Sol. Comparing the given straight line with $\,\,ax^2+2hxy+by^2+2gx+2fy+c=0\,\,$ 

We get, $a=8,\,\,h=5,\,\,b=3,\,\,g=13,\,\,f=8,\,\,c=21.$

# Now, $\,\,\Delta=\begin{vmatrix} a&h &g \\ h& b&f \\ g& f & c \end{vmatrix}=\begin{vmatrix} 8& 5& 13\\ 5& 3 &8 \\ 13& 8 & 21 \end{vmatrix}=0. $

Now, $\,\, 8x^2+(10y+26)x+(3y^2+16y+21)=0 \\ \Rightarrow x=\frac{-(10y+26) \pm \sqrt{(10y+26)^2-4.8.(3y^2+16y+21)}}{2.8}\\=\frac{-(5y+13) \pm \sqrt{(y+1)^2}}{8}\\= \frac{-5y-13+y+1}{8}\\=\frac{-y-3}{2} \longrightarrow(1)\\ \text{and}\,\, x=\frac{-5y-13-(y+1)}{8}=\frac{-3y-7}{4} \longrightarrow (2)\\ $ 

Solving (1) and (2), we get the point of intersection $\,\,(-1,-1).$

Let gradient of (1): $\,m=-2\,$            

and  gradient of (2): $\,m'=-\frac{4}{3}.$                       

So, $ \tan{\theta} \\ = |\frac{-2+4/3}{1+8/3}|=\frac{2}{11}\\ \Rightarrow \theta= \tan^{-1}(2/11)$

# 27. Find the value of 'k' so that $\,\,kx^2-3xy-2y^2+x+13y-15=0\,\,$ represent a pair of straight line. Sol. Comparing the given equation with $ \,\, ax^2+2hxy+by^2+2gx+2fy+c=0\,\,$ we get, $\, a=k,\,h=-3/2,\,b=-2,\,g=1/2,\,f=13/2, \\ \,c=-15\, \\ \text{Now,}\,\, \Delta= \begin{vmatrix} k&-3/2 &1/2 \\ -3/2& -2 & 13/2\\ 1/2 & 13/2 &-15 \end{vmatrix}\\=\frac{-49}{4}k +\frac{196}{8}\longrightarrow (1)\\$ So, by (1), we get $\,\,\Delta =0 \\ \Rightarrow \frac{-49}{4}k +\frac{196}{8}=0 \\ \Rightarrow k=2$

28. Find the equation of the lines passing through $\,\,(2,3)\,\,$ and perpendicular to the lines $\,\,2x^2+xy-3y^2+3x+2y+1=0.$

Sol. $\,\,2x^2+xy-3y^2+3x+2y+1 \\=(l_1x+m_1y+n_1).(l_2x+m_2y+n_2) \\ \therefore l_1l_2=2,\,\,m_1m_2=-3,\,\,l_1m_2+l_2m_1=1, \\ \,n_1n_2=1,\,\,l_1n_2+l_2n_1=3,\,\, \\ m_1n_2+m_2n_1=2 .\\$

Now, equation of lines perpendicular to $\,\,l_1x+m_1y+n_1=0\,\,\text{and}\,\,l_2x+m_2y+n_2=0\,\,\text{are}\,\, \\  m_1x-l_1y+k=0 \longrightarrow(1) \text{and}\\ m_2x-l_2y+k'=0 \longrightarrow(2), \,$ respectively.  

Since (1) and (2) are passing through $\,\,(2,3)\,\,$  we get, $2m_1-3l_1+k=0 \Rightarrow k=3l_1-2m_1 \,\,$  and    $\,\, 2m_2-3l_2+k'=0 \Rightarrow k'=3l_2-2m_2 \\$                                      So, after putting the values of $\,\,k,k'\,$ we get the combined eqn. of (1) and (2), 

$(m_1x-l_1y+3l_1-2m_1) \times  \\ (m_2x-l_2y+3l_2-2m_2)=0 \\ \Rightarrow m_1m_2. x^2 -m_1l_2. xy +3m_1l_2.x -2m_1m_2.x \\-m_2l_1.xy+l_1l_2.y^2-3l_1l_2.y+2l_1m_2y\\+3l_1m_2x-3l_1l_2y+9l_1l_2-6l_1m_2\\-2m_1m_2x+2m_1l_2y-6l_2m_1+4m_1m_2=0\\ \Rightarrow -3x^2-xy+2y^2+3x-12y+12x\\+18-6+2y-12=0 \\ \Rightarrow -3x^2-xy+2y^2+15x-10y=0 \\ \Rightarrow 3x^2+xy-2y^2-15x+10y=0$

29. Find the equation of the sphere which passes through the circle $\,\,x^2+y^2=4;\,z=0 \,\,$ and is not cut by the plane $\,\,x+2y+2z=0\,\,$ in a circle of radius $\,3.\quad \textbf{I.A.S.-2016}$

Sol. The equation of the sphere $\,\, x^2+y^2-4+\lambda z=0 \longrightarrow (1).\\ $        

Comparing (1) with $\,\,x^2+y^2+z^2+2ux+2vy+2wz+d=0,\,\,$ we get              

$\,\, u=0, \, v=0, \, w= \frac{\lambda}{2},\,\,d=-4.\\ \therefore OA= \frac{0+0-2(-\lambda /2)}{\sqrt{1^2+2^2+2^2}}=\frac{\lambda}{3},\\ AB=3,\,\, OB=\sqrt{0^2+0^2+(\frac{\lambda^2}{4})+4} \\=\frac{1}{2}. \sqrt{\lambda^2+4}\,\text{unit}$

Now , $\,\, OA^2+AB^2=OB^2 \\ \Rightarrow \frac{\lambda^2}{9}+9=\frac{1}{4}. (\lambda^2+16) \\ \Rightarrow \lambda= \pm 6.$

 Hence,  the eqn. of the sphere is given by  $\,\, x^2+y^2-4 \pm 6z=0$