If the pair of lines \[6x^2-pxy-3y^2-24x+3y+q=0\] intersect on x- axis then p is equal to :- \[a)3/2\] \[b)-5/2\] \[c)-18\] \[d)-7\]
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Parag Zode
3 years, 7 months ago
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Top NewestFor to intersect this pair of straight line on x-axis , Let They intersect at point P (\(\alpha \, 0\) )
So putting y=0 in given equation
\[6\alpha ^{ 2 }\quad -\quad 24\alpha \quad +\quad q\quad =\quad 0\].
Now since Intersection point is unique so value of \(\alpha \) is also unique. So Discriminant of this quadratic equation is zero !!
\[q\quad =\quad 24\\ \\ \alpha \quad =\quad 2\].
Therefore Given equation representing pair of straight lines if it's Discriminant is zero !!
\[\left| \begin{matrix} 6 & -p/2 & -12 \\ -p/2 & -3 & 3/2 \\ -12 & 3 & 24 \end{matrix} \right| \quad =\quad 0\\ \\ P\quad =\quad 3/2\].
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Ok! I didn't knew to use determinant after finding q and a(alpha).. Anyways ,Nice solution.
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The line(s) intersect the x-axis when y=0. The equation then becomes 6x^2 - 24x + q = 0. There are tons of values for x that will solve this equation depending on the value of q.
If y = 0 then values for p are limitless (infinite).
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Sir but can we assume that for finding value of q the equation\[6x^2-24x+q=0\] should be perfect square ? If it is then the value of q is equal to 24. This is a JEE Question .
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Are there 2 lines? I see only 1 equation.
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The equation above is multiplication of those 2 lines
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What are the 2 lines?
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That is what we have to find in terms of 'p'
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You also have a q which affects the situation.
If this line intersects the x axis then the value for y = 0.
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