Maths adv question (1 Viewer)

[safal.regmi]

This appeared in my trial paper and I still have no idea how to do it, can someone please explain how to do it?

 

[eduiwqh23]

Hell nah

 

[safal.regmi]

Please tell me its actually hard, istg if this is supposed to be easy

 

[eduiwqh23]

Ring up your teacher

 

[Luukas.2]

Any line through the origin has equation for some value of .

This line intersects the curve when .

If the line is a tangent, then the quadratic will have only one solution, and so the discriminant is zero. Thus:

​

and so the two tangents are and .

 

[safal.regmi]

thanks sm.

 

[DoubleDashMan_og]

Ok, so like i'm late to the thread but anyway have a look

so like y=mx+b but the b is zero since the line goes through the origin.

and... this line touches y = x^2+3x+5 twice it's a little cheeky...

so I equate the y's ;

As such, mx=x^2+3x+5

solve for x , 0 = x^2 +x(3-m) + 5

quadratic formula to solve for x but the trick is that the discriminant i.e sqrt(b^2-4ac) = 0 because it's cheeky i.e only touches once

then a wee bit of algebra --> (3-m)^2-4x1x5=0

m= 3+2sqr(5), 3-2sqrt(5)

and then sub into y=mx

Alternative;

dy/dx = 2x+3

dy/dx = m

m=2x+3

mx=x^2+3x+5

sub; (2x+3)=x^2+3x+5

x^2=5

x=+-sqrt(5)

... (therefore soz idk how to do the normal therefore dots) y = +-2sqrt(5)+3 i.e sub x into m=2x+3

soz for weird format lmao